Question 15 Marks
The gravitational attraction between electron and proton in a hydrogen atom is weaker than the coulomb attraction by a factor of about $10^{–40}$. An alternative way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and proton were bound by gravitational attraction. You will find the answer interesting.
AnswerRadius of the first Bohr orbit is given by the relation,
$\text{r}_1=\frac{4\pi\in_0\Big(\frac{\text{h}}{2\pi}\Big)^2}{\text{m}_\text{e}\text{e}^2}\ \dots(1)$
Where,
$\in_0 =$ Permittivity of free space
$h =$ Planck ' s constant $= 6.63 \times 10^{-34} Js$
$me =$ Mass of an electron $= 9.1 \times 10^{-31}kg$
$e =$ Charge of an electron $= 1.9 \times 10^{-19} C$
$mp =$ Mass of a proton $= 1.67 \times 10^{-27} \ kg$
$r =$ Distance between the electron and the proton
Coulomb attraction between an electron and a proton is given as:
$\text{F}_\text{C}=\frac{\text{e}^2}{4\pi\in_0\text{r}^2}\ \dots(2)$
Gravitational force of attraction between an electron and a proton is given as:
$\text{F}_\text{G}=\frac{\text{Gm}_\text{p}\text{m}_\text{c}}{\text{r}^2}\ \dots(3)$
Where,
$G =$ Gravitational constant $= 6.67 \times 10^{-11} N m^{2 }/ \ kg^2$
If the electrostatic $($Coulomb$)$ force and the gravitational force between an electron and a proton are equal, then we can write:
$\therefore\ \text{FG}=\text{FC}$
$\frac{\text{Gm}_\text{p}\text{m}_\text{c}}{\text{r}^2}=\frac{\text{e}^2}{4\pi\in_0\text{r}^2}$
$\therefore\ \frac{\text{e}^2}{4\pi\in_0}=\text{Gm}_\text{p}\text{m}_\text{c}\ \dots(4)$
Putting the value of equation $(4)$ in equation $(1),$ we get:
$\text{r}_1=\frac{\Big(\frac{\text{h}}{2\pi}\Big)^2}{\text{Gm}_\text{p}\text{m}_\text{e}^2}$
$=\frac{\Big(\frac{6.63\times10^{-34}}{2\times3.14}\Big)^2}{6.67\times10^{-11}\times1.67\times10^{-27}\times(9.1\times10^{-31})^2}\approx1.21\times10^{29}\text{ m}$
It is known that the universe is $156$ billion light years wide or $1.5 \times 10^{27} m$ wide.
Hence, we can conclude that the radius of the first Bohr orbit is much greater than the estimated size of the whole universe.
View full question & answer→Question 25 Marks
$A 12.5 eV$ electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?
AnswerIt is given that the energy of the electron beam used to bombard gaseous hydrogen at room temperature is $12.5 eV.$
Also, the energy of the gaseous hydrogen in its ground state at room temperature is $-13.6 ev$.
When gaseous hydrogen is bombarded with an electron beam, the energy of the gaseous hydrogen becomes $-13.6 + 12.5 eV$ i.e., $-1.1 eV.$
Orbital energy is related to orbit level (n) as:
$\text{E}=\frac{-13.6}{(\text{n})^2}\text{ eV}$
$\text{E}=\frac{-13.6}{9}=-1.5\text{ eV}$
For $n = 3,$
This energy is approximately equal to the energy of gaseous hydrogen. It can be concluded that the electron has jumped from $n = 1 $ to $n = 3$ level.
During its de-excitation, the electrons can jump from $n = 3$ to $n = 1$ directly, which forms a line of the Lyman series of the hydrogen spectrum.
We have the relation for wave number for Lyman series as:
$\frac{1}{\lambda}=\text{R}_\text{y}\Big(\frac{1}{1^2}-\frac{1}{\text{n}^2}\Big)$
Where,
$R_y =$ Rydberg constant $= 1.097 x 10^7 rn^{-1}$
$\lambda\ $Wavelength of radiation emitted by the transition of the electron
For $n = 3, $ we can obtain $\lambda$ as:
$\frac{1}{\lambda}=1.097\times10^7\Big(\frac{1}{1^2}-\frac{1}{3^2}\Big)$
$=1.097\times10^7\Big(1-\frac{1}{9}\Big)=1.097\times10^7\times\frac{8}{9}$
$\lambda=\frac{9}{8\times1.097\times10^7}=102.55\text{ nm}$
If the electron jumps from $n = 2$ to $n = 1,$ then the wavelength of the radiation Is given as:
$\frac{1}{\lambda}=1.097\times10^7\Big(\frac{1}{1^2}-\frac{1}{2^2}\Big)$
$=1.097\times10^7\Big(1-\frac{1}{4}\Big)=1.097\times10^7\times\frac{3}{4}$
$\lambda=\frac{4}{1.097\times10^7\times3}=121.54\text{ nm}$
If the transition takes place from $n = 3$ to $n = 2,$ then the wavelength of the radiation is given as:
$\frac{1}{\lambda}=1.097\times10^7\Big(\frac{1}{2^2}-\frac{1}{3^2}\Big)$
$=1.097\times10^7\Big(\frac{1}{4}-\frac{1}{9}\Big)=1.097\times10^7\times\frac{5}{36}$
$\lambda=\frac{36}{5\times1.097\times10^7}=656.33\text{ nm}$
This radiation corresponds to the Balmer series of the hydrogen spectrum.
Hence, in Lyman series, two wavelengths I.e., $102.5 \ nm$ and $121.S$ nm are emitted.
And
in the Balmer series, one wavelength $t.e., 656.33\ nm$ is emitted.
View full question & answer→Question 35 Marks
Obtain an expression for the frequency of radiation emitted when a hydrogen atom de $-$ excites from level n to level $(\text{n – ncy })$ of revolution of the electron in the orbit.
AnswerIt is given that a hydrogen atom de $-$ excites from an upper level $(n)$ to a lower level $(n - 1).$
We have the relation for energy $(E1)$ of radiation at level $n$ as:
$\text{E}_1=\text{hv}_1\frac{\text{hme}^4}{(4\pi)^3\in_0^2\Big(\frac{\text{h}}{2\pi}\Big)}\times\Big(\frac{1}{\text{n}^2}\Big)\ \dots\text(\text{i})$
Where,
$v_1 =$ Frequency of radiation at level $n$
$h =$ plank’s constant
$m =$ mass of hydrogen atom
$e =$ charge on an electron
$\in_0 =$ Permittivity of free space
Now, the relation for energy $(E_2)$ of radiation at level $(n - 1)$ is given as:
$\text{E}_2=\text{hv}_2\frac{\text{hme}^4}{(4\pi)^3\in_0^2\Big(\frac{\text{h}}{2\pi}\Big)}\times\frac{1}{(\text{n}-1)^2}\ \dots\text{(ii)}$
Where,
$v_2 =$ Frequency of radiation at level $(n - 1)$
Energy $(E)$ released as a result of de $-$ excitation:
$E = E_2 -E_1$
$hv = E_2 -E_{1 }...(iii)$
Where,
$v =$ Frequency of radiation emitted
Putting values from equations $(i)$ and $(ii)$ in equation $(iii),$ we get:
$\text{v}=\frac{\text{me}^4}{(4\pi)^3\in^2_0\Big(\frac{\text{h}}{2\pi}\Big)^3}\Big[\frac{1}{(\text{n}-1)^2}-\frac{1}{\text{n}^2}\Big]$
$=\frac{\text{me}^4(2\text{n}-1)}{(4\pi)^3\in^2_0\Big(\frac{\text{h}}{2\pi}\Big)^3\text{n}^2(\text{n}-1)^2}$
For large $n,$ we can write $(2\text{n}-1)\approx2\text{n}$ and $(\text{n}-1)\approx\text{n}$
$\therefore\ \text{v}=\frac{\text{me}^4}{32\pi\in^2_0\Big(\frac{\text{h}}{2\pi}\Big)^3\text{n}^3}\ \dots(\text{iv)}$
Classical relation of frequency of revolution of an electron is given as:
$\text{v}_\text{c}=\frac{\text{v}}{2\pi\text{r}}\ \dots\text{(v})$
Where,
Velocity of the electron in the nth orbit is given as:
$\text{v}=\frac{\text{e}^2}{4\pi\in_0\Big(\frac{\text{h}}{2\pi}\Big)\text{n}}\ \dots(\text{vi)}$
And, radius of the nth orbit is given as:
$\text{r}=\frac{4\pi\in_0\Big(\frac{\text{h}}{2\pi}\Big)^2}{\text{me}^2}\text{ n}^2\ \dots\text{(vii)}$
Putting the values of equations $(vi)$ and $(vii)$ in equation $(v),$ we get:
$\text{v}_\text{c}=\frac{\text{me}^4}{32\pi^3\in^2_0\Big(\frac{\text{h}}{2\pi}\Big)^3\text{n}^3}\ \dots\text{(viii)}$
Hence, the frequency of radiation emitted by the hydrogen atom is equal to its classical orbital frequency.
View full question & answer→Question 45 Marks
A hydrogen atom initially in the ground level absorbs a photon, which excites it to the $n = 4$ level. Determine the wavelength and frequency of photon.
AnswerFor ground level $, n_1 = 1$
Let $E_1$ be the energy of this level. It ls known that $E_1$ Is related with $n_1$ as:
$\text{E}_1=\frac{-13.6}{\text{n}^2_1}\text{ eV}$
$=\frac{-13.6}{1^2}=-13.6\text{ eV}$
The atom is excited to a higher level $, n_2 = 4.$
Let $E_2$ be the energy of this level.
$\therefore\ \text{E}_2=\frac{-13.6}{\text{n}^2_2}\text{ eV}$
$=\frac{-13.6}{4^2}=-\frac{13.6}{16}\text{ eV}$
The amount of energy absorbed by the photon is given as:
$E = E_2 - E_1$
$=\frac{-13.6}{16}-\Big(-\frac{13.6}{1}\Big)$
$=\frac{13.6\times15}{16}\text{ eV}$
$=\frac{13.6\times15}{16}\times1.6\times10^{-19}=2.04\times10^{-18}\text{ J}$
For a photon of wavelenqtha. the expression of energy Is written as:
$\text{E}=\frac{\text{hc}}{\lambda}$
Where,
$h =$ Planck's constant $= 6.6 x 10^{-34} Js$
$c =$ Speed of light $= 3 x 10^8m/s$
$\therefore\ \lambda=\frac{\text{hc}}{\text{E}}$
$=\frac{6.6\times10^{-34}\times3\times10^8}{2.04\times10^{-18}}$
$=9.7\times10^{-8}\text{ m}=97\text{ nm}$
And, frequency of a photon is given by the relation,
$\text{v}=\frac{\text{c}}{\lambda}$
$=\frac{3\times10^8}{9.710^{-8}}\approx3.1\times10^{15}\text{ Hz}$
Hence, the wavelength of the photon is $97 \ nm$ while the frequency is $3.1 x 10^{15} Hz.$
View full question & answer→Question 55 Marks
- Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the $n = 1, 2,$ and $3$ levels.
- Calculate the orbital period in each of these levels.
Answer
- Let $v_1$ be the orbital speed of the electron in a hydrogen atom in the ground state level $, n_1 = 1$. For charge $(e)$ of an electron, $v,$ is given by the relation,
$\text{v}_1\frac{\text{e}^2}{\text{n}_14\pi\in_0(\text{h}/2\pi)}=\frac{\text{e}^2}{2\in_0\text{h}}$
Where,
$e = 1.6 \times 10^{-19} C$
$\in_0$ = Permittivity of free space $= 8.85\times10^{-12}\text{ N}^{-1}\text{ C}^2\text{ m}^{-2}$
$h =$ Planck's constant $= 6.62 x 10^{-34} Js$
$\therefore\ \text{v}_1=\frac{\big(1.6\times10^{-19}\big)^2}{2\times8.85\times10^{-12}\times6.62\times10^{-34}}$
$= 0.0218 \times 10^8 = 2.18 \times 10^6 m/s$
For level $n_2^= 2,$ we can write the relation for the corresponding orbital speed as:
$\text{v}_1=\frac{\text{e}^2}{\text{n}_22\in_0\text{h}}$
$=\frac{\big(1.6\times10^{-19}\big)^2}{2\times2\times8.85\times10^{-12}\times6.62\times10^{-34}}$
$= 1.09 \times 10^6 m/s$
And, for $n_3 = 3,$ we can write the relation for the corresponding orbital speed as:
$\text{v}_3=\frac{\text{e}^3}{\text{n}_22\in_0\text{h}}$
$=\frac{\big(1.6\times10^{-19}\big)^2}{3\times2\times8.85\times10^{-12}\times6.62\times10^{-34}}$
$= 7.27 \times 10^5 m/s$
Hence, the speed of the electron in a hydrogen atom in $n = 1, n = 2$ and $n = 3$ is $2.18 \times 10^6 m/s, 1.09 \times 10^6 m/s, 7.27 \times 10^5 m/s$ respectively.
- Let $T_1$ be the orbital period of the electron when it is in level $n_1 = 1$. Orbital period is related to orbital speed as:
$\text{T}_1=\frac{2\pi\text{r}_1}{\text{v}_1}$
Where,
$r_1=$ Radius of the orbit
$=\frac{\text{n}^2_1\text{h}^2\in_0}{\pi\text{me}^2}$
$h =$ Planck's constant $= 6.62 x 10^{-34} Js$
$e =$ Charge on an electron $= 1.6 x 10^{-19} C$
$\in_0 = $ Permittivity of free space $= 8.85 x 10^{-12} N^{-1} C^2m^{-2}$
$m =$ Mass of an electron $= 9.1 x 10^{-31} \ kg$
$\therefore\ \text{T}_1=\frac{2\pi\text{r}_1}{\text{v}_1}$
$=\frac{2\pi\times(1)^2\times\big(6.62\times10^{-34}\big)^2\times8.85\times10^{-12}}{2.18\times10^6\times\pi\times9.1\times10^{-31}\times\big(1.6\times10^{-19}\big)^1}$
$= 15.27 \times 10^{-17} = 1.527 \times 10^{-16} s$
For level $n_2 = 2,$ we can write the period as:
$\therefore\ \text{T}_2=\frac{2\pi\text{r}_2}{\text{v}_2}$
Where,
$r_2 =$ Radius of the electron in $n_2 = 2$
$=\frac{(\text{n}_2)^2\text{h}^2\in_0}{\pi\text{me}^2}$
$\therefore\ \text{T}_2=\frac{2\pi\text{r}_2}{\text{v}_2}$
$=\frac{2\pi\times(2)^2\times\big(6.62\times10^{-34}\big)^2\times8.85\times10^{-12}}{1.09\times10^6\times\pi\times9.1\times10^{-31}\times\big(1.6\times10^{-19}\big)^2}$
$= 1.22 \times 10^{-15} s$
And, for level $n_3 = 3,$ we can write the period as:
$\therefore\ \text{T}_3=\frac{2\pi\text{r}_3}{\text{v}_3}$
Where,
$r_3=$ Radius of the electron in $n_3 = 3$
$=\frac{(\text{n}_3)^2\text{h}^2\in_0}{\pi\text{me}^2}$
$\therefore\ \text{T}_3=\frac{2\pi\text{r}_3}{\text{v}_3}$
$=\frac{2\pi\times(3)^2\times\big(6.62\times10^{-34}\big)^2\times8.85\times10^{-12}}{7.27\times10^5\times\pi\times9.1\times10^{-31}\times\big(1.6\times10^{-19}\big)^2}$
$= 4.12 \times 10^{-15} s$
Hence, the orbital period in each of these levels is $1.52 x 10^{-16} s, 1.22 x 10^{-15 }s,$ and $4.12 \times 10-^{15} s$ respectively. View full question & answer→Question 65 Marks
Classically, an electron can be in any orbit around the nucleus of an atom. Then what determines the typical atomic size? Why is an atom not, say, thousand times bigger than its typical size? The question had greatly puzzled Bohr before he arrived at his famous model of the atom that you have learnt in the text. To simulate what he might well have done before his discovery, let us play as follows with the basic constants of nature and see if we can get a quantity with the dimensions of length that is roughly equal to the known size of an atom $(~ 10^{–10}m).$
- Construct a quantity with the dimensions of length from the fundamental constants $e, m_e,$ and $c$. Determine its numerical value.
- You will find that the length obtained in $(a)$ is many orders of magnitude smaller than the atomic dimensions. Further, it involves $c$. But energies of atoms are mostly in non $-$ relativistic domain where c is not expected to play any role. This is what may have suggested Bohr to discard c and look for ‘something else’ to get the right atomic size. Now, the Planck’s constant h had already made its appearance elsewhere. Bohr’s great insight lay in recognising that $h, me, $ and e will yield the right atomic size. Construct a quantity with the dimension of length from $h, m_e,$ and e and confirm that its numerical value has indeed the correct order of magnitude.
Answer
- Charge on an electron $, e = 1.6 \times 10^{-19} C$
Mass of an electron $, me = 9.1 \times 10^{-31} \ kg$
Speed of light $, c = 3 \times 10^8 m/s$
Let us take a quantity involving the given quantities as $\Bigg(\frac{\text{e}^2}{4\pi\in_0\text{m}_\text{e}\text{C}^2}\Bigg)$
Where,
$\in_0 =$ Permittivity of free space
And, $\frac{1}{4\pi\in_0}=9\times10^9\text{Nm}^2\text{C}^{-2}$
The numerical value of the taken quantity will be:
$\frac{1}{4\pi\in_0}\times\frac{\text{e}^2}{\text{m}_\text{e}\text{c}^2}$
$=9\times10^9\times\frac{(1.6\times10^{-19})^2}{9.1\times10^{-31}\times(3\times10^8)^2}$
$=2.81\times10^{-15}\text{ m}$
Hence, the numerical value of the taken quantity is much smaller than the typical size of an atom.
- Charge on an electron $e = 1.6 \times 10^{-19}C$
Mass of an electron $, me = 9.1 \times 10^{-31} \ kg$
Planck’s constant $, h = 6.63 \times 10^{-34}Js$
Let us take a quantity involving the given quantities as $\Bigg(\frac{\text{e}^2}{4\pi\in^0\text{m}_\text{e}\text{C}^2}\Bigg)$
Where,
$\in_0 =$ Permittivity of free space
And, $\frac{1}{4\pi\in_0}=9\times10^9\text{Nm}^2\text{C}^{-2}$
The numerical value of the taken quantity will be:
$\frac{1}{4\pi\in_0}\times\frac{(1.6\times10^{-10})^2}{9.1\times10^{-31}\times(3\times10^8)^2}$
$=2.81\times10^{-15}\text{ m}$
Hence, the value of the quantity taken is of the order of the atomic size. View full question & answer→Question 75 Marks
In a Geiger-Marsden experiment, what is the distance of closest approach to the nucleus of a $7.7 MeV \alpha$-particle before it comes momentarily to rest and reverses its direction?
AnswerThe key idea here is that throughout the scattering process, the total mechanical energy of the system consisting of an $\alpha$-particle and a gold nucleus is conserved. The system's initial mechanical energy is $E_i$, before the particle and nucleus interact, and it is equal to its mechanical energy $E_f$ when the $\alpha$-particle momentarily stops. The initial energy $E_i$ is just the kinetic energy $K$ of the incoming $\alpha$ - particle. The final energy $E_f$ is just the electric potential energy $U$ of the system. The potential energy $U$ can be calculated from Eq. (12.1).
Let $d$ be the centre-to-centre distance between the $\alpha$-particle and the gold nucleus when the $\alpha$-particle is at its stopping point. Then we can write the conservation of energy $E_i=E_f$ as
$
K=\frac{1}{4 \pi \varepsilon_0} \frac{(2 e)(Z e)}{d}=\frac{2 Z e^2}{4 \pi \varepsilon_0 d}
$
Thus the distance of closest approach $d$ is given by
$
d=\frac{2 Z e^2}{4 \pi \varepsilon_0 K}
$
The maximum kinetic energy found in $\alpha$-particles of natural origin is $7.7 MeV$ or $1.2 \times 10^{-12} J$. Since $1 / 4 \pi \varepsilon_0=9.0 \times 10^9 N m ^2 / C ^2$. Therefore with $e=1.6 \times 10^{-19} C$, we have,
$
\begin{aligned}
d & =\frac{(2)\left(9.0 \times 10^9 Nm ^2 / C^2\right)\left(1.6 \times 10^{-19} C \right)^2 Z }{1.2 \times 10^{-12} J } \\
& =3.84 \times 10^{-16} Zm
\end{aligned}
$
The atomic number of foil material gold is $Z=79$, so that $d( Au )=3.0 \times 10^{-14} m =30 fm$. $(1 fm$ (i.e. fermi $)=10^{-15} m$. $)$
The radius of gold nucleus is, therefore, less than $3.0 \times 10^{-14} m$. This is not in very good agreement with the observed result as the actual radius of gold nucleus is $6 fm$. The cause of discrepancy is that the distance of closest approach is considerably larger than the sum of the radii of the gold nucleus and the $\alpha$-particle. Thus, the $\alpha$-particle reverses its motion without ever actually touching the gold nucleus.
View full question & answer→Question 85 Marks
In the Auger process an atom makes a transition to a lower state without emitting a photon. The excess energy is transferred to an outer electron which may be ejected by the atom. (This is called an Auger electron). Assuming the nucleus to be massive, calculate the kinetic energy of an n = 4 Auger electron emitted by Chromium by absorbing the energy from a n = 2 to n = 1 transition.
AnswerAs the nucleus is massive, recoil momentum of the atom may be neglected and the entire energy of the transition may be considered transferred to the Auger electron. As there is a single valence electron in Cr, the energy states may be thought of as given by the Bohr model.The energy of the nth state $\text{En}=-\text{Z}^2\text{R}\frac{1}{\text{n}^2}$ where R is the Rydberg constant and Z = 24.
The energy released in a transition form 2 to 1 is $\Delta\text{t}=\text{Z}^2\text{R}\Big(1-\frac{1}{4}\Big)=\frac{3}{4}\text{Z}^2\text{R}$.
The energy required to eject a n = 4 electron is $\text{E}_4=\text{Z}^2\text{R}\frac{1}{16}$.
Thus, the kinetic energy of the Auger electron is $\text{KE}=\text{Z}^2\text{R}\Big(\frac{3}{4}-\frac{1}{16}\Big)=\frac{1}{16}\text{Z}^2\text{R}$
$=\frac{11}{16}\times24\times24\times13.6\text{eV}=5385.6\text{eV}.$
View full question & answer→Question 95 Marks
We have stimulated emission and spontaneous emission. Do we also have stimulated absorption and spontaneous absorption?
AnswerWhen a photon of energy $(E_2 - E_1 = hυ)$ is incident on an atom in the ground state, the atom in the ground state $E_1$ may absorb the photon and jump to a higher energy state $(E_2).$ This process is called stimulated absorption or induced absorption.Spontaneous absorption is the process by which an atom in its ground state spontaneously jumps to a higher energy state, resulting in the absorption of a photon. We do not have any process such as spontaneous absorption. This is because for absorption, we need to incident a photon of sufficient energy on the atom to stimulate the atom for absorption.
View full question & answer→Question 105 Marks
Show that the first few frequencies of light that is emitted when electrons fall to the $n^{th}$ level from levels higher than $n,$ are approximate harmonics $($i.e. in the ratio $1 : 2: 3...)$ when $n > > 1$.
AnswerKey concept:Frequency of emitted radiation:
$\Delta\text{E}=\text{hf}$
$\Rightarrow\ \text{f}=\frac{\Delta\text{E}}{\text{h}}=\frac{\text{E}_2-\text{E}_1}{\text{h}}=\text{RcZ}^2\bigg(\frac{1}{\text{n}_1^2}-\frac{1}{\text{n}_2^1}\bigg)$
Wave number/wavelength:
Wave number is the number of waves in unit length
$\text{v}=\frac{1}{\lambda}=\frac{\text{f}}{\text{c}}$
$\Rightarrow\ \frac{1}{\lambda}=\text{RZ}^2\bigg(\frac{1}{\text{n}_1^2}-\frac{1}{\text{n}_2^2}\bigg)=\frac{13.6\text{Z}^2}{\text{hc}}\bigg(\frac{1}{\text{n}_1^2}-\frac{1}{\text{n}_2^2}\bigg)$
Number of spectral lines: If an electron jumps from higher energy orbit to lower energy orbit it emits radiations with various spectral lines.
If electron falls from orbit $n_2$ to $n_1$ then the number of spectral lines emitted is given by
$\text{N}_\text{E}=\frac{(\text{n}_2-\text{n}_1+1)(\text{n}_2-\text{n}_1)}{2}$
If electron falls from $n^{th}$ orbit to ground state $($i.e. $n_2 = n$ and $n_1 = 1),$ then number of spectral lines emitted $\text{N}_\text{E}=\frac{\text{n}(\text{n}-1)}{2}$
The frequency of any line in a series in the spectrum of hydrogen like atoms correspnding to the transiton of electrons from $(n + p)$ level to $n^{th}$ level can be expressed as a difference of two terms;
$\text{f}_\text{mn}=\text{cRZ}^2\bigg[\frac{1}{(\text{n}+\text{p})^2}-\frac{1}{\text{n}^2}\bigg]$
where $, m = n + p, (p = 1, 2, 3, ....)$ and $R$ is Rydberg constant.
For $p < < n$
$\text{f}_\text{mn}=\text{cRZ}^2\bigg[\frac{1}{\text{n}^2}\Big(1+\frac{\text{p}}{\text{n}}\Big)^{-2}-\frac{1}{\text{n}^2}\bigg]$
$\text{f}_\text{mn}=\text{cRZ}^2\bigg[\frac{1}{\text{n}^2}-\frac{2\text{p}}{\text{n}}-\frac{1}{\text{n}^2}\bigg] \ [$By Binomial theorem $(1 + x)n = 1 + nx$ if $|x| < 1]$
$\text{f}_\text{mn}=\text{cRZ}^2\frac{2\text{p}}{\text{n}^3}\simeq\bigg(\frac{2\text{cRZ}^2}{\text{n}^3}\bigg)\text{p}$
Hence, the first few frequencies of light that is emitted when electrons fall to the nth level from levels higher that n, are approximately harmonic $($i.e., in the ratio $1 : 2 : 3 ....)$ when $n > > 1.$
View full question & answer→Question 115 Marks
Derive the expression for the magnetic field at the site of a point nucleus in a hydrogen atom due to the circular motion of the electron. Assume that the atom is in its ground state and give the answer in terms of fundamental constants.
AnswerTo keep the electron in its orbit, the centripetal force on the electron must be equal to the electrostatic force of attraction. Therefore,$\frac{\text{mv}^2}{\text{r}}=\frac{1}{4\pi\varepsilon_0}\frac{\text{e}^2}{\text{r}^2}\ ...(\text{i})$ (For H atom, Z = 1)
From bohr's quantisation condition
$\text{mvr}=\frac{\text{nh}}{2\pi}\Rightarrow\text{v}=\frac{\text{nh}}{2\pi\text{mr}}$
For K shell, n = 1
$\text{v}=\frac{\text{nh}}{2\pi\text{mr}} \ ...(\text{ii})$
From (i) and (ii), we have
$\frac{\text{m}}{\text{r}}\Big(\frac{\text{h}}{2\pi\text{mr}}\Big)^2=\frac{1}{4\pi\varepsilon_0}\frac{\text{e}^2}{\text{r}^2}$
$\frac{\text{m}}{\text{r}}\frac{\text{h}^2}{4\pi^2\text{m}^2\text{r}^2}=\frac{1}{4\pi\varepsilon_0}\frac{\text{e}^2}{\text{r}^2}\Rightarrow\pi\text{rme}^2=\varepsilon_0\text{h}^2$
$\text{r}=\frac{\varepsilon_0\text{h}^2}{\pi\text{me}^2} \ ...(\text{iii})$
From (ii) and (iii), we have
$\text{v}=\frac{\text{h}\times\pi\text{me}^2}{2\pi\text{m}\varepsilon_0\text{h}^2}=\frac{\text{e}^2}{2\varepsilon_0\text{h}}$
Magnetic field at the centre of a circular loop $\text{B}=\frac{\pi_0\text{I}}{2\text{r}}$
$\text{I}=\frac{\text{Charge}}{\text{Time}}$ and $\text{Time}=\frac{2\pi\text{r}}{\text{v}}$
$\therefore\text{I}=\frac{\text{ev}}{2\pi\text{r}}$
So, $\text{B}=\frac{\mu_0\text{ev}}{2\text{r}\times2\pi\text{r}}=\frac{\mu_0\text{ev}}{4\pi\text{r}^2} \ ...(\text{iv})$
From (ii), (iii), (iv), we have
$\text{B}=\frac{\pi_0\text{e.e}^2\pi^2\text{m}^2\text{e}^4}{2\varepsilon_0\text{h}\times4\pi\times\varepsilon^2_0\text{h}^4}\Rightarrow\text{B}=\frac{\mu_0\text{e}^7\pi\text{m}^2}{8\varepsilon^3_0\text{h}^5}$
View full question & answer→Question 125 Marks
Consider an excited hydrogen atom in state $n$ moving with a velocity $u(ν << c)$. It emits a photon in the direction of its motion and changes its state to a lower state $m$. Apply momentum and energy conservation principles to calculate the frequency $ν$ of the emitted radiation. Compare this with the frequency $ν_0$ emitted if the atom were at rest.
AnswerVelocity of hydrogen atom in state $‘n’ = u$
Also the velocity of photon $= u$
But $u << C$
Here the photon is emitted as a wave.
So its velocity is same as that of hydrogen atom i.e. $u$.
$\therefore$ Accounding to Doppler’s effect,
Frequency $\text{v}=\text{v}_0\bigg(\frac{1+\frac{\text{u}}{\text{c}}}{1-\frac{\text{u}}{\text{c}}}\bigg)$
as $u <<< C$
$1-\frac{\text{u}}{\text{c}}=\text{q}$
$\therefore\ \text{v}=\text{v}_0\bigg(\frac{1+\frac{\text{u}}{\text{c}}}{1}\bigg)=\text{v}_0\Big(1+\frac{\text{u}}{\text{c}}\Big)$
$\text{v}=\text{v}_0\Big(1+\frac{\text{u}}{\text{c}}\Big)$
View full question & answer→Question 135 Marks
A beam of monochromatic light of wavelength $\lambda$ ejects photoelectrons from a cesium surface $\big(\Phi=1.9\text{eV}\big).$ These photoelectrons are made to collide with hydrogen atoms in ground state. Find the maximum value of $\lambda$ for which (a) hydrogen atoms may be ionized, (b) hydrogen atoms may get excited from the ground state to the first excited state and (c) the excited hydrogen atoms may emit visible light.
Answer$\phi=1.9\text{eV}$
- The hydrogen is ionized,
$\text{n}_1=1,\text{n}_2=\infty$
Energy required for ionization $=13.6\bigg(\frac{1}{\text{n}^2_1}-\frac{1}{\text{n}^2_1}\bigg)=13.6$
$\frac{\text{hc}}{\lambda}-1.9=13.6$
$\lambda=80.1\text{nm}=80\text{nm}$
- For the electron to be excited from,
$\text{n}_1\text{ to n}_2=2$
$\text{E}=13.6\bigg(\frac{1}{\text{n}^2_1}-\frac{1}{\text{n}^2_1}\bigg)=13.6\Big(1-\frac{1}{4}\Big)=\frac{13.6\times3}{4}$
$\frac{\text{hc}}{\lambda}-1.9=\frac{13.6\times3}{4}$
$\lambda=\frac{1242}{12.1}=102.64=102\text{nm}$
- $\text{E}_2=13.6\bigg(\frac{1}{\text{n}^2_1}-\frac{1}{\text{n}^2_1}\bigg)$
$\text{E}_2=13.6\Big(\frac{1}{4}-\frac{1}{9}\Big)$
$\text{E}_2=\frac{13.66\times5}{36}$
For Einstein's photoelecrtic equation,
$\frac{\text{hc}}{\lambda}-1.9=\frac{13.6\times5}{36}$
$\frac{\text{hc}}{\lambda}=\frac{13.6\times5}{36}+1.9$
$\frac{1240}{\lambda1}=1.88+1.9=3.78$
$\lambda=\frac{1240}{3.78}$
$\lambda=328.04\text{nm}$ View full question & answer→Question 145 Marks
Assume that there is no repulsive force between the electrons in an atom but the force between positive and negative charges is given by Coulomb’s law as usual. Under such circumstances, calculate the ground state energy of a $He -$ atom.
AnswerKey concept: Total energy $($for Hydrogen and $H_2$ like Atoms$)$: The total energy of the electron in the $n^{th}$ stationary states of the hydrogen.
Atom of hydrogen $-$ like atom of atomic number $Z$ is given by,
$\text{E}=-\Big(\frac{\text{me}^4}{8\epsilon_0\text{h}^2}\Big).\frac{\text{Z}^2}{\text{n}^2}=-\Big(\frac{\text{me}^2}{8\epsilon_0\text{ch}^3}\Big)\text{ch}\frac{\text{Z}^2}{\text{n}^2}$
$=-\text{R ch}\frac{\text{Z}^2}{\text{n}^2}=-13.6\frac{\text{Z}^2}{\text{n}^2}\text{eV}$
For a $He -$ nucleus $Z = 2,$ and for ground state $n = 1.$
Thus, ground state energy of a $He -$ atom.
$\text{E}_\text{n}=-13.6\frac{\text{Z}^2}{\text{n}^2}\text{eV}=-13.6\frac{2^2}{1^2}\text{eV}=-54.5\text{eV}$
Thus, the ground state will have two electrons each of energy $E$ and the total ground state energy would be $-(4 \times 13.6)eV = 54.5eV.$
View full question & answer→Question 155 Marks
The inverse square law in electrostatics is $|\text{F}|=\frac{\text{e}^2}{(4\pi\epsilon_0)\text{r}^2}$ for the force between an electron and a proton. The $\Big(\frac{1}{\text{r}}\Big)$ dependence of $|F|$ can be understood in quantum theory as being due to the fact that the 'particle' of light $($photon$)$ is massless. If photons had a mass $m_p$, force would be modified to $|\text{F}|=\frac{\text{e}^2}{(4\pi\epsilon_0)\text{r}^2}\Big[\frac{1}{\text{r}^2}+\frac{\lambda}{\text{r}}\Big].\text{e}\times\text{p}(-\lambda\text{r})$ where $\lambda=\frac{\text{m}_\text{p}\text{c}}{\text{h}}$ and $\text{h}=\frac{\text{h}}{2\pi}$. Estimate the change in the ground state energy of a $H-$ atom if $m_p$ were $10^{-6}$ times the mass of an electron.
AnswerWe are given $\lambda=\frac{\text{m}_\text{p}\text{c}}{\text{h}}=\frac{\text{m}_\text{p}\text{c}^2}{\text{hc}}=\frac{(10^{-6}\text{m}_\text{e})\text{c}^2}{\text{hc}}$
$=\frac{10^{-6}[0.51](1.6\times10^{-13}\text{J})3\times10^8\text{ms}^{-1}}{(1.05\times10^{-34}\text{Js})(3\times10^8\text{ms}^{-1})}$
$=0.26\times10^7\text{m}^{-1},\ \ [\because\ \text{m}_\text{e}\text{c}^2=0.51\text{MeV}]$
$\text{r}_\text{B}(\text{Bohr's radius}=051\mathring{\text{A}}=0.51\times10^{-10}\text{m}$
or $\lambda\text{r}_\text{B}=(0.26\times10^7\text{m}^{-1})(0.51\times10^{-10}\text{m})$
$=0.14\times10^{-3}< < 1$
Furhter, as $|\text{F}|=\Big(\frac{\text{e}^2}{4\pi\epsilon_0}\Big)\bigg[\frac{1}{\text{r}^2}+\frac{\lambda}{\text{r}}\bigg]\text{e}^{\lambda\text{r}}\ .....(\text{i})$
and $|\text{F}|=\frac{\text{dU}}{\text{dr}},$
$\text{U}_\text{r}=\int|\text{F}|\text{dr}=\Big(\frac{\text{e}^2}{4\pi\epsilon_0}\Big)\int\Big(\frac{\lambda_\text{e}^{-\lambda\text{r}}}{\text{r}}+\frac{\text{e}^{-\lambda\text{r}}}{\text{r}^2}\Big)\text{dr}$
If $\text{z}=\frac{\text{e}^{-\lambda\text{r}}}{\text{r}}=\frac{1}{\text{r}}(\text{e}^{-\lambda\text{r}}),$
$\frac{\text{dz}}{\text{dr}}=\Big[\frac{1}{\text{r}}(\text{e}^{-\lambda\text{r}})(-\lambda)+(\text{e}^{-\lambda\text{r}})\Big(-\frac{1}{\text{r}^2}\Big)\Big]$
or $\text{dz}=-\Big[\frac{\lambda\text{e}^{-\lambda\text{r}}}{\text{r}}+\frac{\text{e}^{-\lambda\text{r}}}{\text{r}^2}\Big]\text{dr}$
Thus $\int\Big(\frac{\lambda\text{e}^{-\lambda\text{r}}}{\text{r}}+\frac{\text{e}^{-\lambda\text{r}}}{\text{r}^2}\Big)\text{dr}$
$\Rightarrow-\int\text{dz}=-\text{z}=-\frac{\text{e}^{-\lambda\text{r}}}{\text{r}}$
$=-\Big(\frac{\text{e}^2}{4\pi\text{E}_0}\Big)\Big(\frac{\text{e}^{-\lambda\text{r}}}{\text{r}}\Big)\ .....(\text{ii})$
We know that
$\text{mvr}=\text{h}$
$\Rightarrow\ \text{v}=\frac{\text{h}}{\text{mr}};\text{ and }$
$\frac{\text{mv}^2}{\text{r}}=\text{F}=\Big(\frac{\text{e}^2}{4\pi\epsilon_0}\Big)\Big(\frac{1}{\text{r}^2}+\frac{\lambda}{\text{r}}\Big)$
$[$ Putting $\text{e}^{-\lambda\text{r}}\approx1$ in eqn. $(i)]$
Thus, $\Big(\frac{\text{m}}{\text{r}}\Big)\Big(\frac{\text{h}^2}{\text{m}^2\text{r}^2}\Big)=\Big(\frac{\text{e}^2}{4\pi\epsilon_0}\Big)\Big(\frac{1}{\text{r}^2}+\frac{\lambda}{\text{r}}\Big)$
or $\frac{\text{h}^2}{\text{m}}=\Big(\frac{\text{e}^2}{4\pi\epsilon_0}\Big)\Big(\frac{\text{r}+\lambda\text{r}^2}{\text{r}^3}\Big)$
or $\frac{\text{h}^2}{\text{m}}=\Big(\frac{\text{e}^2}{4\pi\epsilon_0}\Big)({\text{r}+\lambda\text{r}^2})\ .....(\text{iii})$
When $\lambda=0,\text{r}=\text{r}'_\text{B}$ and
$\frac{\text{h}^2}{\text{m}}=\Big(\frac{\text{e}^2}{4\pi\epsilon_0}\Big)\text{r}_\text{B}\ .....(\text{iv})$
From eqns. $(iii)$ and $(iv)$,
$\text{r}_\text{B}+\text{r}+\lambda\text{r}^2$
Let $\text{r}=\text{r}_\text{B}+\delta$ so that from $(iii)$
$\text{r}_\text{B}=(\text{r}_\text{B}+\delta)+\lambda(\text{r}^2_\text{B}+\delta^2+2\delta\text{r}_\text{B})$
or $0=\lambda\text{r}_\text{B}^2+\delta(1+2\lambda\text{r}_\text{B})\ \ (\text{neglecting }\delta^2)$
or $\delta=-\frac{\lambda\text{r}_\text{B}^2}{(1+2\lambda\text{r}_\text{B})}=(-\lambda\text{r}_\text{B}^2)(1+2\lambda_\text{B})^{-1}$
$=(-\lambda\text{r}_\text{B}^2)(1+2\lambda_\text{B})^{-1}=-\lambda\text{r}_\text{B}^2\ \ (\because\ \lambda\text{r}_\text{B}< < 1)$
From eqn. $ (ii) \text{u}_\text{r}=-\Big(\frac{\text{e}^2}{4\pi\epsilon_0}\Big)\frac{\text{e}^{-\lambda(\text{r}_\text{B}+\delta)}}{(\text{r}_\text{B}+\delta)}$
$=-\Big(\frac{\text{e}^2}{4\pi\epsilon_0}\frac{1}{\text{r}_\text{B}}\Big)\Big(1-\frac{\delta}{\text{r}_\text{B}}\Big)(1-\lambda\text{r}_\text{B})\approx-\frac{\text{e}^2}{4\pi\epsilon_0\text{r}_\text{B}}$
$=-24.2\text{eV}$
$[\because\ \text{e}^{-\lambda(\text{r}_\text{B}+\delta)}\approx\ 1-\lambda(\text{r}_\text{B}+\delta)=1-\lambda\text{r}_\text{B}-\lambda\delta\approx1-\lambda\text{e}_\text{B}]$
and $\frac{1}{(\text{r}_\text{B}+\delta)}=\frac{1}{\text{r}_\text{B}(1+\frac{\delta}{\text{r}_\text{B}})}=\frac{1}{\text{r}_\text{B}}\Big(1+\frac{\delta}{\text{r}_\text{B}}\Big)^{-1}$
$=\frac{1}{\text{r}_\text{B}}\Big(1-\frac{\delta}{\text{r}_\text{B}}\Big)$
Further $, KE$ of the electron,
$\text{K}=\frac{1}{2}\text{mv}^2=\frac{1}{2}\text{m}\Big(\frac{\text{h}^2}{\text{m}^2\text{r}^2}\Big)$
$=\frac{\text{h}^2}{2\text{mr}^2}=\frac{\text{h}^2}{2\text{m}(\text{r}_\text{B}+\delta)^2}=\frac{\text{h}^2}{2\text{mr}_\text{B}^2+(1+\frac{\delta}{\text{r}_\text{B}})^2}$
$\Big(\frac{\text{h}^2}{2\text{mr}^2_\text{B}}\Big)\Big(1+\frac{\delta}{\text{r}_\text{B}}\Big)^{-2}=\Big(\frac{\text{h}^2}{2\text{mr}^2_\text{B}}\Big)\Big(1-\frac{\delta}{\text{r}_\text{B}}\Big)$
$=(13.6)(1+2\lambda\text{r}_\text{B})\text{eV}$
$\Big(\text{as}\frac{\text{h}^2}{2\text{mr}^2_\text{B}}=13.6\text{eV and }\delta=-\lambda\text{r}^2_\text{B}\Big)$
Total energy of $H-$ atom in the ground state $=$ final energy $-$ initial energy
$=(-13.6+27.2\lambda\text{r}_\text{B})\text{eV}-(-13.6\text{eV})$
$=(27.2\lambda\text{r}_\text{B})\text{eV}$
View full question & answer→Question 165 Marks
What is the minimum energy that must be given to a $H$ atom in ground state so that it can emit an $\text{H}_\gamma$ line in Balmer series. If the angular momentum of the system is conserved, what would be the angular momentum of such $\text{H}_\gamma$ photon?
Answer$\text{H}_\gamma$ in Balmer series corresponds to transition $n = 5$ to $n = 2.$
So, the electron in ground state $n = 1$ must first be put in state $n = 5.$
Energy required $= E_1 - E_5 = 13.6 - 0.54 = 13.06\ eV$
If angular momentum is conserved, angular momentun of photon $=$ change in angular momentum of electron $= L_5 - L_2 $
$= 5h - 2h $
$= 3 \times 1.06 \times 10^{-34} $
$= 3.18 \times 10^{-34}$
$ = 3.18 \times 10^{-38}kh \ m^2/s.$
View full question & answer→Question 175 Marks
What will be the energy corresponding to the first excited state of a hydrogen atom if the potential energy of the atom is taken to be $10eV$ when the electron is widely separated from the proton? Can we still write $\text{E}_\text{n}=\frac{\text{E}_1}{\text{n}^2}?\text{ r}_\text{n}=\text{a}_0\text{n}^2?$
AnswerEnergy of $n^{th}$ state of hydrogen is given by,
$\text{E}_\text{n}=\frac{-13.6}{\text{n}^2}\text{eV}$
Energy of first excited state $(n = 2)$ of hydrogen, $\text{E}_\text{n}=\frac{-13.6}{\text{n}^2}\text{eV}=-3.4\text{eV}$
This relation holds true when the refrence point energy is zero.
Usually the refrence point energy is the energy of the atom when the electron is widely separated from the proton. In the given question, the potential energy of the atom is taken to be $10eV$ when the electron is widely separated from the proton so here our refrence point energy is $10eV$.
Earlier,
The energy of first excited state was $-3.4eV$ when the refrence point had zero energy but now as the refrence point has shifted so
The energy of the first excited state will also shift by the corresponding amount.
Thus,
$\text{E}_1=-3.4\text{eV}-10\text{eV}=-13.4\text{eV}$
We still write $\text{E}_\text{n}=\frac{\text{E}_1}{\text{n}^2},\text{ or }\text{r}_\text{n}=\text{a}_0\text{n}^2$ because these formulas are independent of the refrence point energy.
View full question & answer→Question 185 Marks
The earth revolves round the sun due to gravitational attraction. Suppose that the sun and the earth are point particles with their existing masses and that Bohr's quantization rule for angular momentum is valid in the case of gravitation.
(a) Calculate the minimum radius the earth can have for its orbit.
(b) What is the value of the principal quantum number n for the present radius? Mass of the earth $= 6.0 \times 10^{-24}kg.$ Mass of the sun $= 2.0 \times 10^{30}kg,$ earth $-$ sun distance $= 1.5 \times 10^{11}m$.
AnswerMass of Earth $= Me = 6.0 \times 1024\ kg$ Mass of Sun $= Ms = 2.0 \times 1030\ k$g Earth $-$ Sun dist $= 1.5 \times 1011\ m \text{mvr}=\frac{\text{nh}}{2\pi}\text{ or m}^2\text{v}^2\text{r}^2=\frac{\text{n}^2\text{h}^2}{4\pi^2}\ ....(\text{i})$
$\frac{\text{GMeMs}}{\text{r}^2}=\frac{\text{mev}^2}{\text{r}}\text{ or v}^2=\frac{\text{GM}}{\text{s/r}}\ ....(\text{ii})$ Dividing $(i)$ and $(ii),$
We get $\text{me}^2\text{r}=\frac{\text{n}^2\text{h}^2}{4\pi^2\text{GMs}}$
- For $n = 1$
$\text{r}=\sqrt{\frac{\text{h}^2}{4\pi^2\text{GMsMe}^2}}=2.29\times10^{-138}\text{m}$
$\text{r}=2.3\times10^{-138}\text{m}$
- $\text{n}^2=\frac{\text{Me}^2\times\text{r}\times4\times\pi^2\times\text{G}\times\text{Ms}}{\text{h}^2}$
$\text{n}^2=2.5\times10^{74}$ View full question & answer→Question 195 Marks
An atom is in its excited state. Does the probability of its coming to ground state depend on whether the radiation is already present or not? If yes, does it also depend on the wavelength of the radiation present?
AnswerWhen an atom transits from an excited state to ground state in the presence of an external radiation then it is called as stimulated transition.
When an atom transits from an excited state to ground state on its own then it is called as spontaneous transition.
Ratio of the coefficient for stimulated transition to spontaneous transition is given by,
$\text{R}=\frac{\text{A}}{\text{B}\rho(\nu)}=\text{e}\frac{\text{h}\nu}{\text{kT}}-1$
For microwave region
$\nu=10^{10}(\text{say})$
$\text{R}=\text{e}^{\frac{6.6\times10^{-34}\times10^{10}}{1.38\times10^{-23}\times300}}-1$
$\text{R}=\text{e}^{0.0016}-1$
$\text{R}=0.0016$
This implies that stimulated transition dominate in this region.
For visible region,
$\nu=10^{15}$
$\text{R}=\text{e}^{160}-1$
$\text{R}>>1$
So here spontaneous transition dominate.
View full question & answer→Question 205 Marks
Positronium is just like a $H-$ atom with the proton replaced by the positively charged anti $-$ particle of the electron $($called the positron which is as massive as the electron$)$. What would be the ground state energy of positronium?
AnswerKey concept: Positronium $(Ps)$ is a system consisting of an electron and its anti $-$ particle a positron, bound together into an exotic atom, specifically anonium.
The system is unstable: the two particles annihilate each other to predominantly produce two or three gamma $-$ rays, depending on the relative spin states.
The orbit and energy levels of the two particles arc similar to that of the hydrogen atom $($which is a bound slate of a proton and an electron$)$. $I$
lowever, because of the reduced mass, the frequencies of the spectral lines are less than half of the corresponding hydrogen lines.
The total energy of the electron in the starionary states of the hydrogen atom is given by
$\text{E}_\text{n}=-\left\{\frac{\mu}{2\text{h}^2}\Big(\frac{\text{e}^2}{4\pi\epsilon_0}\Big)^2\right\}\frac{1}{\text{n}^2}$
$\mu$ that occure in the Bohr formula is the reduced mass of electron and proton.
For hydrogen atom: $\mu=\frac{\text{m}_\text{e}\text{m}_\text{p}}{\text{m}_\text{e}+\text{m}_\text{p}}\approx\frac{\text{m}_\text{e}\text{m}_\text{p}}{\text{m}_\text{p}}=\text{m}_\text{e}(\text{ As m}_\text{p}>>\text{m}_\text{e})$
$m_e, m_p $ are the mass of electron and proton which are the same.
For positronium: $\mu=\frac{\text{m}_\text{e}\text{m}_\text{p}}{\text{m}_\text{e}+\text{m}_\text{p}}=\frac{\text{m}_\text{e}^2}{2\text{m}_\text{e}}=\frac{\text{m}_\text{e}}{2}$
$($As mass of positron is equal to the mass of electron$)$
$\text{E}_\text{n}=-\left\{\frac{\mu}{2\text{h}^2}\Big(\frac{\text{e}^2}{4\pi\epsilon_0}\Big)^2\right\}\frac{1}{\text{n}^2}=-\frac{13.6\text{eV}}{\text{n}^2}=\frac{-6.8\text{eV}}{\text{n}^2}$
The energy is half of the hydrogen level.
The lowest energy of positronium $(n = 1)$ is $-6.8$ electron volts $(eV)$.
The next highest energy level $(n = 2) $ is $-1.7 eV,$ the negative sign implies a bound state.
View full question & answer→Question 215 Marks
If a proton had a radius $R$ and the charge was uniformly distributed, calculate using Bohr theory, the ground state energy of a $H-$ atom when:
- $R = 0.1 \mathring A .$
- $R = 10 \mathring A .$
AnswerFor a point nucleus in $H-$ atom,
Ground state: $mvr = h,$
$\frac{\text{mv}^2}{\text{r}_\text{B}}=-\frac{\text{e}^2}{\text{r}^2_\text{B}}.\frac{1}{4\pi\epsilon_0}$
The electrostatic force of attraction between positively charged nucleus and negatively charged electrons $($Coulombian force$)$ provides necessary centripetal force of revolution.
$\frac{\text{mv}^2}{\text{r}_\text{B}}=-\frac{\text{e}_2}{\text{r}^2_\text{B}}.\frac{1}{4\pi\epsilon_0}$
By Bohr's postulates in ground state, we have
$mvr = h$
$\therefore\ \text{m}\frac{\text{h}^2}{\text{m}^2\text{r}^2_\text{B}}=+\Big(\frac{\text{e}^2}{4\pi\epsilon_0}\Big)\frac{1}{\text{r}_\text{B}^2}$
$\therefore\ \frac{\text{h}^2}{\text{m}}.\frac{4\pi\epsilon_0}{\text{e}^2}=\text{r}_\text{B}=0.51\mathring{\text{A}}$
Potential energy
$-\Big(\frac{\text{e}^2}{4\pi\text{r}_0}\Big).\frac{1}{\text{r}_\text{B}}=-27.2\text{ev}; \text{KE}=\frac{\text{mv}^2}{2}=\frac{1}{2}\text{m}.\frac{\text{h}^2}{\text{m}^2\text{r}^2_\text{B}}=\frac{\text{h}}{2\text{mr}^2_\text{B}}=+13.6\text{eV}$
For a spherical nucleus of radius $R,$
If $R < r_B,$ same result.
If $R > > r_B$ the electron moves inside the sphere with redius $r'_B (r'_B=$ new Bohr radius$)$.
Charge inside, $\text{r}_\text{B}^4=\text{e}\Big(\frac{\text{r}^3_\text{B}}{\text{R}^3}\Big)$
$\therefore\ \text{r}'_\text{B}=\frac{\text{h}^2}{\text{m}}\Big(\frac{4\pi\epsilon_0}{\text{e}^2}\Big)\frac{\text{R}^3}{\text{r}_\text{B}'^3}$
$\text{r}_\text{B}'^4=(0.51\mathring{\text{A}}),\text{R}^3\ \ [\text{R}=10\mathring{\text{A}}]$
$=510(\mathring{\text{A}})^4$
$\therefore\ \text{r}'\text{B}\approx(510)^{\frac{1}{4}}\mathring{\text{A}}<\text{R}$
$\text{KE}=\frac{1}{2}\text{mv}^2=\frac{\text{m}}{2}.\frac{\text{h}}{\text{m}^2\text{r}'^2_\text{B}}=\frac{\text{h}}{2\text{m}}.\frac{1}{\text{r}'^2_\text{B}}$
$=\Big(\frac{\text{h}}{2\text{mr}^2_\text{B}}\Big)\Big(\frac{\text{r}^2_\text{B}}{\text{r}'^2_\text{B}}\Big)=(13.6\text{eV})\frac{(0.51)^2}{(510)^\frac{1}{2}}=\frac{3.54}{22.6}=0.16\text{eV}$
$\text{PE}=+\Big(\frac{\text{e}^2}{4\pi\epsilon_0}\Big).\Big(\frac{\text{r}'^2-3\text{R}^2}{2\text{R}^3}\Big)=+\Big(\frac{\text{e}^2}{4\pi\epsilon_0}.\frac{1}{\text{r}_\text{B}}\Big)\bigg(\frac{\text{r}_\text{B}(\text{r}'^2_\text{B}-3\text{R}^2)}{\text{R}^3}\bigg)$
$=+(27.2\text{eV})\bigg[\frac{0.51(\sqrt{510}-300)}{1000}\bigg]$
$=+(27.2\text{eV}),\frac{-141}{1000}=-3.83\text{eV}$
View full question & answer→Question 225 Marks
Deutrium was discovered in $1932$ by Harold Urey by measuring the small change in wavelength for a particular transition in $1H$ and $2H$. This is because, the wavelength of transition depend to a certain extent on the nuclear mass. If nuclear motion is taken into account then the electrons and nucleus revolve around their common centre of mass. Such a system is equivalent to a single particle with a reduced mass $\mu ,$ revolving around the nucleus at a distance equal to the electron $-$ nucleus separation. Here $\mu=\frac{\text{m}_\text{e}\text{M}}{\text{m}_\text{e}+\text{M}}$ where $M$ is the nuclear mass and me is the electronic mass. Estimate the percentage difference in wavelength for the $1^{st}$ line of the Lyman series in $1H$ and $2H. ($Mass of $1H$ nucleus is $1.6725 \times 10^{-27}kg,$ Mass of $2H$ nucleus is $3.3374 \times 10^{-27}kg$, Mass of electron $= 9.109 \times 10^{-31}kg).$
AnswerAccording to the Bohr's theory for the hydrogen like atom of atomic number $Z,$ the total energy of the electron in the $n^{th}$ state is $\text{E}_\text{n}=-\left\{\frac{\mu}{2\text{h}^2}\Big(\frac{\text{e}^2}{4\pi\epsilon_0}\Big)^2\right\}\frac{\text{Z}^2}{\text{n}^2}=-\frac{\mu\text{Z}^2\text{e}^4}{8\epsilon_0^2\text{h}^2}\Big(\frac{1}{\text{h}^2}\Big)$ where signs are as usual and the $\mu$ that occurs in the Bohr formula is the reduced mass of electron and proton. For hydrogen atom: $\mu_\text{H}=\frac{\text{m}_\text{e}\text{M}_\text{H}}{(\text{m}_\text{e}+\text{M}_\text{H})}$ For Deutrium: $\mu_\text{D}=\frac{\text{m}_\text{e}\text{M}_\text{D}}{(\text{m}_\text{e}+\text{M}_\text{D})}$ Let $\mu_\text{H}$ be the reduced mass o hydrogen is $\text{hf}_\text{H}=\frac{\mu\text{Z}^2\text{e}^4}{8\epsilon_0^2\text{h}^2}\Big(1-\frac{1}{4}\Big)=\frac{\mu_\text{H}\text{e}^4}{8\epsilon_0^2\text{h}^2}\times\frac{3}{4}$ Thus, the wavelength of the transition is $\lambda\text{H}=\frac{3}{4}\frac{\mu_\text{H}\text{e}^4}{8\epsilon_0^2\text{h}^3\text{c}}$. The wavelength of the treansition for the same line in Deutrium is $\lambda_\text{D}=\frac{3}{4}\frac{\mu_\text{D}\text{e}^4}{8\epsilon_0^2\text{h}^3\text{c}}$. $\therefore$ Difference in wavelength $\Delta\lambda=\lambda_\text{D}-\lambda_\text{H}$ Hence, the percentage difference is $100\times\frac{\Delta\lambda}{\lambda_\text{H}}=\frac{\lambda_\text{D}-\lambda_\text{H}}{\lambda_\text{H}}\times100=\frac{\mu_0-\mu_\text{H}}{\mu_\text{H}}\times100$
$=\frac{\frac{\text{m}_\text{e}\text{M}_\text{D}}{(\text{m}_\text{e}+\text{M}_\text{D})}-\frac{\text{m}_\text{e}\text{M}_\text{H}}{(\text{m}_\text{e}+\text{M}_\text{H})}}{\frac{\text{m}_\text{e}\text{M}_\text{H}}{(\text{m}_\text{e}+\text{M}_\text{H})}}\times100$
$=\bigg[\Big(\frac{\text{m}_\text{e}+\text{M}_\text{H}}{\text{m}_\text{e}+\text{M}_\text{D}}\Big)\frac{\text{M}_\text{D}}{\text{M}_\text{H}}-1\bigg]\times100$
Sicne $, m_e < < M_H < < M_D \frac{\Delta\lambda}{\lambda_\text{H}}\times100=\begin{bmatrix} \frac{\text{M}_\text{H}}{\text{M}_\text{H}}\times\frac{\text{M}_\text{D}}{\text{M}_\text{H}}\Bigg(\frac{1+\frac{\text{m}_\text{e}}{\text{M}_\text{H}}}{1+\frac{\text{m}_\text{e}}{\text{M}_\text{D}}}\Bigg)-1 \end{bmatrix}\times100$
$=\bigg[\Big(1+\frac{\text{m}_\text{e}}{\text{M}_\text{H}}\Big)\Big(1+\frac{\text{m}_\text{e}}{\text{M}_\text{D}}\Big)^{-1}-1\bigg]\times100$
$=\bigg[\Big(1+\frac{\text{m}_\text{e}}{\text{M}_\text{H}}\Big)\Big(1+\frac{\text{m}_\text{e}}{\text{M}_\text{D}}\Big)-1\bigg]\times100$
$[$By binomial theorem $, (1 + x)n = 1 + nx is |x| < 1]$
$\frac{\Delta\lambda}{\lambda_\text{H}}\times100\bigg[1+\frac{\text{m}_\text{e}}{\text{M}_\text{H}}-\frac{\text{m}_\text{e}}{\text{M}_\text{D}}-\frac{(\text{m}_\text{e})^2}{{\text{M}_\text{H}}{\text{M}_\text{D}}}-1\bigg]\times100$
Nwglecting $\frac{(\text{m}_\text{e})^2}{\text{M}_\text{H}\text{M}_\text{D}}$, as it is very small.
$\Rightarrow\ \frac{\Delta\lambda}{\lambda_\text{H}}\times100\simeq\Big[\frac{\text{m}_\text{e}}{\text{M}_\text{H}}-\frac{\text{m}_\text{e}}{\text{M}_\text{D}}\Big]\times100$
$\Rightarrow\ \frac{\Delta\lambda}{\lambda_\text{H}}\times100\approx\Big[\frac{1}{\text{M}_\text{H}}-\frac{1}{\text{M}_\text{D}}\Big]\times100$
$=9.1\times10^{-31}\bigg[\frac{1}{1.6725\times10^{-27}}-\frac{1}{3.3374\times10^{-27}}\bigg]\times100$
$=9.1\times10^{-4}[0.5979-0.2996]\times100=2.714\times10^{-2}\%$
View full question & answer→Question 235 Marks
How many wavelengths are emitted by atomic hydrogen in visible range $(380\ nm-780\ nm)$? In the range $50\ nm$ to $100\ nm$?
AnswerBalmer series contains wavelengths ranging from $364\ nm \ ($for $n_2 = 3)$ to $655\text{ nm}(\text{n}_2)=\infty$
So, the given range of wavelength $(380-780\ nm)$ lies in the Balmer series.
The wavelength in the Balmer series can be found by,
$\frac{1}{\lambda}=\text{R}\Big(\frac{1}{2^2}-\frac{1}{\text{n}^2}\Big)$
Here $, R =$ Rydberg's constant $= 1097 \times 107m^{-1}$
The wavelength for the transition from $n = 3$ to $n = 2$ is given by,
$\frac{1}{\lambda_1}=\text{R}\Big(\frac{1}{2^2}-\frac{1}{\text{3}^2}\Big)$
$\lambda_1=656.3\text{ nm}$
The wavelength for the transition from $n = 4$ to $n = 2$ is given by,
$\frac{1}{\lambda_2}=\text{R}\Big(\frac{1}{2^2}-\frac{1}{\text{4}^2}\Big)$
$\lambda_2=486.1\text{ nm}$
The wavelength for the transition from $n = 5$ to $n = 2$ is given by,
$\frac{1}{\lambda_3}=\text{R}\Big(\frac{1}{2^2}-\frac{1}{\text{5}^2}\Big)$
$\lambda_3=434.0\text{ nm}$
The wavelength for the transition from $n = 6$ to $n = 2$ is given by,
$\frac{1}{\lambda_4}=\text{R}\Big(\frac{1}{2^2}-\frac{1}{\text{6}^2}\Big)$
$\lambda_4=410.2\text{ nm}$
The eavelength for the transition from $n = 7$ to $n = 2$ is given by,
$\frac{1}{\lambda_5}=\text{R}\Big(\frac{1}{2^2}-\frac{1}{\text{7}^2}\Big)$
$\lambda_5=397.0\text{ nm}$
Thus, the wavelengths emitted by the atomic hydrogen in visible range $(380-780\ nm)$ are $5$.
Lyman series contains wavelengths ranging from $91\ nm \ ($for $n_2 = 2)$ to $121\text{ nm}(\text{n}_2)=\infty$
So, the wavelengths in the given range $(50-100\ nm)$ must lie in the Lyman series.
The wavelength in the Lyman series can be found by,
$\frac{1}{\lambda}=\text{R}\Big(\frac{1}{1^2}-\frac{1}{\text{n}^2}\Big)$
The wavelength for the transition from $n = 2$ to $n = 1$ is given by,
$\frac{1}{\lambda_1}=\text{R}\Big(\frac{1}{1^2}-\frac{1}{2^2}\Big)$
$\lambda_1=122\text{ nm}$
The wavelength for the transition from $n = 3$ to $n = 1$ is given by,
$\frac{1}{\lambda_2}=\text{R}\Big(\frac{1}{1^2}-\frac{1}{3^2}\Big)$
$\lambda_2=103\text{ nm}$
The wavelength for the transition from $n = 4$ to $n = 1$ is given by,
$\frac{1}{\lambda_3}=\text{R}\Big(\frac{1}{1^2}-\frac{1}{4^2}\Big)$
$\lambda_3=97.3\text{ nm}$
The wevelength for the transion from $n = 5$ to $n = 1$ is given by,
$\frac{1}{\lambda_4}=\text{R}\Big(\frac{1}{1^2}-\frac{1}{5^2}\Big)$
$\lambda_4=95.0\text{ nm}$
The eavelength for the transition from $n = 6$ to $n = 1$ is given by,
$\frac{1}{\lambda_5}=\text{R}\Big(\frac{1}{1^2}-\frac{1}{6^2}\Big)$
$\lambda_5=93.8\text{ nm}$
So, it can be noted that the number of wevelengths laying between $50\ nm$ to $100\ nm$ are $3$
View full question & answer→Question 245 Marks
A parallel beam of light of wavelength $100nm$ passes through a sample of atomic hydrogen gas in ground state. $(a)$ Assume that when a photon supplies some of its energy to a hydrogen atom, the rest of the energy appears as another photon. Neglecting the light emitted by the excited hydrogen atoms in the direction of the incident beam, what wavelengths may be observed in the transmitted beam? $(b)$ A radiation detector is placed near the gas to detect radiation coming perpendicular to the incident beam. Find the wavelengths of radiation that may be detected by the detector.
Answer$\lambda=100\text{ nm}$
$\text{E}=\frac{\text{hc}}{\lambda}=\frac{1242}{100}=12.42\text{eV}$
- The possible transitions may be $E_1$ to $E_2$
$E_1$ to $E_2,$ energy absorbed $= 10.2eV$
Energy left $= 12.42 - 10.2 = 2.22eV$
$2.22\text{eV}=\frac{\text{hc}}{\lambda}=\frac{1242}{\lambda}$ or $\lambda=559.45=560\text{ nm}$
$E_1$ to $E_3,$ Energy absorbed $= 12.1eV$
Energy left $= 12.42 - 12.1 = 0.32eV$
$0.32=\frac{\text{hc}}{\lambda}=\frac{1242}{\lambda}$ or $\lambda=\frac{1242}{0.32}=3881.2=3881\text{ nm}$
$E_3$ to $ E_4,$ Energy absorbed $= 0.65$
Energy left $= 12.42 - 0.65 = 11.77eV$
$11.77=\frac{\text{hc}}{\lambda}=\frac{1242}{\lambda}$ or $\lambda=\frac{1242}{11.77}=105.52$
- The energy absorbed by the $H$ atom is now radiated perpendicular to the incident beam.
$10.2=\frac{\text{hc}}{\lambda}$ or $\lambda=\frac{1242}{10.2}=121.76\text{ nm}$
$12.1=\frac{\text{hc}}{\lambda}$ or $\lambda=\frac{1242}{12.1}=102.64\text{ nm}$
$0.65=\frac{\text{hc}}{\lambda}$ or $\lambda=\frac{1242}{0.65}=1910.76\text{ nm}$ View full question & answer→Question 255 Marks
Consider a neutron and an electron bound to each other due to gravitational force. Assuming Bohr's quantization rule for angular momentum to be valid in this case, derive an expression for the energy of the neutron-electron system.
Answer$\text{m}_\text{e}\text{Vr}=\frac{\text{nh}}{2\pi}\ ...(\text{i})$
$\frac{\text{GM}_\text{n}\text{M}_\text{e}}{\text{r}^2}=\frac{\text{m}_\text{e}\text{V}^2}{\text{r}}$
$\frac{\text{GM}_\text{n}}{\text{r}}=\text{v}^2\ ...(\text{ii})$
Squaring (ii) and dividing it with (i)
$\frac{\text{m}_\text{e}^2\text{v}^2\text{r}^2}{\text{v}^2}=\frac{\text{n}^2\text{h}^2\text{r}}{4\pi^2\text{Gm}_\text{n}}$
$\text{me}^2\text{r}=\frac{\text{n}^2\text{h}^2\text{r}}{4\pi^2\text{Gm}_\text{n}}$
$\text{r}=\frac{\text{n}^2\text{h}^2\text{r}}{4\pi^2\text{Gm}_\text{n}\text{me}^2}$
$\text{v}=\frac{\text{nh}}{2\pi\text{rm}_\text{e}}$ [from (i)]
$\text{v}=\frac{\text{nh4}\pi^2\text{GM}_\text{n}\text{M}^2_\text{e}}{2\pi\text{M}_\text{e}\text{n}^2\text{h}^2}=\frac{2\pi\text{GM}_\text{n}\text{M}_\text{e}}{\text{nh}}$
$\text{KE}=\frac{1}{2}\text{m}_\text{e}\text{V}^2$
$=\frac{1}{2}\text{m}_\text{e}\frac{\big(2\pi\text{GM}_\text{n}\text{M}_\text{e}\big)^2}{\text{nh}}=\frac{4\pi^2\text{G}^2\text{M}^2_\text{n}\text{M}^3_\text{e}}{2\text{n}^2\text{h}^2}$
$\text{PE}=\frac{-\text{GM}_\text{n}\text{M}_\text{e}}{\text{r}}$
$=\frac{-\text{GM}_\text{n}\text{M}_\text{e}4\pi^2\text{GM}_\text{n}\text{M}^2_\text{e}}{\text{n}^2\text{h}^2}=\frac{-4\pi^2\text{G}^2\text{M}^2_\text{n}\text{M}^3_\text{e}}{\text{n}^2\text{h}^2}$
Total energy $=\text{KE}+\text{PE}=\frac{2\pi^2\text{G}^2\text{M}^2_\text{n}\text{M}^3_\text{e}}{2\text{n}^2\text{h}^2}$
View full question & answer→Question 265 Marks
Find the wavelength of the radiation emitted by hydrogen in the transitions $(a) n = 3$ to $n= 2, (b) n = 5$ to $n = 4$ and $(c) n = 10$ to $n = 9.$
AnswerFrom Balmer empirical formula, the wavelength $(\lambda)\lambda$ of the radiation is given by, $\frac{1}{\lambda}=\text{R}\Big(\frac{1}{\text{n}_1^2}-\frac{1}{\text{n}^2_2}\Big)$
Here $, R = $ Rydberg constant $= 1.097 \times 10^7m^{-1} n_1 =$ Quantum number of final state $ n_2 =$ Quantum number of initial state
- For transition from$ n = 3$ to $n = 2$,
Here,$ n_1 = 2, n_2 = 3$
$\frac{1}{\lambda}=1.09737\times10^{7}\times\Big(\frac{1}{4}-\frac{1}{9}\Big)$
$\lambda=\frac{36}{5\ \times\ 1.09737\ \times\ 10^{7}}$
$=6.56\times10^{-7}=656\text{ nm}$
- For transition from $n = 5$ to $n = 4,$
Here, $n_1 = 4, n_2 = 5$
$\frac{1}{\lambda}=1.09737\times10^{-7}\Big(\frac{1}{16}-\frac{1}{25}\Big)$
$\lambda=\frac{400}{1.09737\ \times10^7\ \times\ 9}$
$\lambda=4050\text{ nm}$
- For transition from $n = 10$ to $n = 9,$
Here, $n_1 = 9, n_2 = 10$
$\frac{1}{\lambda}=1.09737\times10^7\Big(\frac{1}{81}-\frac{1}{100}\Big)$
$\lambda=\frac{81\ \times\ 100}{19\ \times\ 1.09737\ \times\ 10^7}$
$\lambda=38849\text{ nm}$ View full question & answer→Question 275 Marks
The first four spectral lines in the Lyman serics of a $H-$ atom are $\lambda = 1218 \mathring A , 1028 \mathring A , 974.3 \mathring A$ and $951.4 \mathring A .$ If instead of Hydrogen, we consider Deuterium, calculate the shift in the wavelength of these lines.
AnswerLet $\mu_H$ and $\mu_D$ are the reduced masses of electron for hydrogen and deuterium respectively.
We know that $\frac{1}{\lambda}=\text{R}\Big[\frac{1}{\text{n}^2_\text{f}}-\frac{1}{\text{n}^2_\text{i}}\Big]$
As ni and nf are fixed for by mass series for hudrogen and deuterium.
$\lambda\propto\frac{1}{\text{r}}$ or $\frac{\lambda_\text{D}}{\lambda_\text{H}}=\frac{\text{R}_\text{H}}{\text{R}_\text{D}}\ .....(\text{i})$
$\text{R}_\text{R}=\frac{\text{m}_\text{e}\text{e}^4}{8\epsilon_0\text{ch}^3}=\frac{\mu_\text{H}\text{e}^4}{8\epsilon_0\text{ch}^3}$
$\text{R}_\text{H}=\frac{\text{m}_\text{e}\text{e}^4}{8\epsilon_0\text{ch}^3}=\frac{\mu_\text{H}\text{e}^4}{8\epsilon_0\text{ch}^3}$
$\therefore\ \frac{\text{R}_\text{H}}{\text{R}_\text{D}}=\frac{\text{R}_\text{H}}{\text{R}_\text{D}}\ .....(\text{ii})$
From equation $(i)$ and $(ii)$
$\frac{\lambda_\text{D}}{\lambda_\text{H}}=\frac{\mu_\text{H}}{\mu_\text{D}}\ .....(\text{iii})$
Reduced mass for hydrogen,
$\mu_\text{H}=\frac{\text{m}_\text{e}}{1+\frac{\text{m}_\text{e}}{\text{m}}}\simeq\text{m}_\text{e}\Big(1-\frac{\text{m}_\text{e}}{2\text{M}}\Big)$
Reduced mass for deuterium,
$\mu_\text{D}=\frac{2\text{M.m}_\text{e}}{2\text{M}\Big(1+\frac{\text{m}_\text{e}}{2\text{M}}\Big)}\simeq\text{m}_\text{e}\Big(1-\frac{\text{m}_\text{e}}{2\text{m}}\Big)$
where $M$ is mass of proton
$\frac{\mu_\text{H}}{\mu_\text{D}}\frac{\text{m}_\text{e}\Big(1-\frac{\text{m}_\text{e}}{2\text{M}}\Big)}{\text{m}_\text{e}\Big(1-\frac{\text{m}_\text{e}}{2\text{M}}\Big)}=\Big(1-\frac{\text{m}_\text{e}}{\text{M}}\Big)\Big(1-\frac{\text{m}_\text{e}}{2\text{M}}\Big)^{-1}$
$=\Big(1-\frac{\text{m}_\text{e}}{\text{M}}\Big)\Big(1+\frac{\text{m}_\text{e}}{2\text{M}}\Big)$
$\Rightarrow\ \frac{\mu_\text{H}}{\mu_\text{D}}=\Big(1-\frac{\text{m}_\text{e}}{2\text{M}}\Big)$
or $\frac{\mu_\text{H}}{\mu_\text{D}}=\Big(1-\frac{1}{2\times1840}\Big)=0.99973\ .....(\text{iv})$
$(\because\ \text{M}=1840\text{m}_\text{e})$
From $(iii)$ and $(iv)$
$\frac{\lambda_\text{D}}{\lambda_\text{H}}=0.99973,\lambda_\text{D}=0.99973\lambda_\text{H}.$
Using $\lambda_\text{H}=1218\mathring{\text{A}},1028\mathring{\text{A}},974.3\mathring{\text{A}}\text{ and }951.4\mathring{\text{A}}$, we get
$\lambda_\text{D}=1217.7\mathring{\text{A}},1027.7\mathring{\text{A}},974.04\mathring{\text{A}},951.1\mathring{\text{A}}$
Shift in wavelength $(\lambda_\text{H}-\lambda_\text{D})\approx0.3\mathring{\text{A}}.$
View full question & answer→Question 285 Marks
A uniform magnetic field B exist in a region. An electron projected perpendicular to the field goes in a circle. Assuming Bohr's quantization rule for angular momentum, calculate (a) the smallest possible radius of the electron (b) the radius of the nth orbit and (c) the minimum possible speed of the electron.
AnswerAccording to Bohr’s quantization rule, $\text{mvr}=\frac{\text{nh}}{2\pi}$ ‘r’ is less when ‘n’ has least value i.e. 1 or, $\text{mv}=\frac{\text{nh}}{2\pi\text{R}}\ ...(\text{i})$ Again, $\text{r}=\frac{\text{mv}}{\text{qB}},\text{ or }\text{mv}=\text{rqB}\ ...(\text{ii})$
- From (i) and (ii)
$\text{rqB}=\frac{\text{nh}}{2\pi\text{r}}$ [q = e]
$\text{r}^2=\frac{\text{nh}}{2\pi\text{eB}}$
$\text{r}=\sqrt{\frac{\text{h}}{2\pi\text{eB}}}$ [here n = 1]
- For the radius of nth orbit, $\text{r}=\sqrt{\frac{\text{nh}}{2\pi\text{eB}}}$
- $\text{mvr}=\frac{\text{nh}}{2\pi},\ \text{r}=\frac{\text{mv}}{\text{qB}}$
Substituting the value of ‘r’ in (i)
$\text{mv}\times\frac{\text{mv}}{\text{qB}}=\frac{\text{nh}}{2\pi}$
$\text{m}^2\text{v}^2=\frac{\text{nheB}}{2\pi\text{m}^2}$ [n = 1, q = e]
$\text{v}^{2}=\frac{\text{heB}}{2\pi\text{m}^2}$
$\text{v}=\sqrt{\frac{\text{heB}}{2\pi\text{m}^2}}$ View full question & answer→Question 295 Marks
A hydrogen atom moving at speed $υ$ collides with another hydrogen atom kept at rest. Find the minimum value of $υ$ for which one of the atoms may get ionized. The mass of a hydrogen atom $= 1.67 \times 10^{-27}kg.$
AnswerThe hydrogen atoms after collision move with speeds $v_1$ and $v_2$
$\text{mv}=\text{mv}_1+\text{mv}_2\ ....(\text{i})$
$\frac{1}{2}\text{mv}^2=\frac{1}{2}\text{mv}^2_1+\frac{1}{2}\text{mv}^2_2+\Delta\text{E}\ ....(\text{ii})$
From $(i), \text{v}^2=(\text{v}_1+\text{v}_2)^2=\text{v}^2_1+\text{v}^2_2+2\text{v}_1\text{v}_2$
From $ (ii), \text{v}^2=\text{v}^2_1+\text{v}^2_2+\frac{2\Delta\text{E}}{\text{m}}$
$=2\text{v}_1\text{v}_2=\frac{2\Delta\text{E}}{\text{m}}\ ...(\text{iii})$
$(\text{v}_1-\text{v}_2)^2=\big(\text{v}_1+\text{v}_2\big)^2-4\text{v}_1\text{v}_2$
$(\text{v}_1-\text{v}_2)=\text{v}^2-\frac{4\Delta\text{E}}{\text{m}}$
For minimum value of $‘v\ ’$
$\text{v}_1=\text{v}_2$
$\text{v}^2-\Big(\frac{4\Delta\text{E}}{\text{m}}\Big)=0$
$\text{v}^2=\frac{4\Delta\text{E}}{\text{m}}=\frac{4\times13.6\times1.6\times10^{-19}}{1.67\times10^{-27}}$
$\text{v}=\sqrt{\frac{4\times13.6\times1.6\times10^{-19}}{1.67\times10^{-27}}}=7.2\times10^4\text{m/s}$
View full question & answer→Question 305 Marks
The Bohr model for the $H-$ atom relies on the Coulomb’s law of electrostatics. Coulomb’s law has not directly been verified for very short distances of the order of angstroms. Supposing Coulomb’s law between two opposite charge $+q_1, -q_2$ is modified to
$|\text{F}|=\frac{\text{q}_1\text{q}_2}{(4\pi\epsilon_0)}\frac{1}{\text{r}^2}.\text{r}\geq\text{R}_0$
$=\frac{\text{q}_1\text{q}_2}{(4\pi\epsilon_0)}\frac{1}{\text{R}^2_0}\Big(\frac{\text{R}_0}{\text{r}}\Big)^{\epsilon}.\text{r}\geq\text{R}_0$
Calculate in such a case, the ground state energy of a $H-$ atom, if $\epsilon=0.1,\text{R}_0=1\mathring{\text{A}}$.
AnswerLeu us consider the case, when $\text{r}\leq\text{R}_0=1.\mathring{\text{A}}$
Let $\epsilon=2+\delta$
$\text{F}=\frac{\text{q}_1\text{q}_2}{4\pi\epsilon_0}.\frac{\text{R}_\delta}{\text{r}^{2+\delta}}=\frac{\text{xR}_0^\delta}{\text{r}^{2+\delta}}$
Where, $\frac{\text{q}_1\text{q}_2}{4\pi_0\epsilon_0}=\text{x}=(1.6\times10^{-19})^2\times9\times10^9$
$=2.04\times10^{-29}\text{Nm}^2$
The electrostatic force of attraction between positively charged nucleus and negatively charged electrons $($Coulombian force$)$ provides necessary centripetal froce.
$\frac{\text{mv}^2}{\text{r}}=\frac{\text{xR}_0^\delta}{\text{r}^{2+\delta}}\text{ or v}^2=\frac{\text{xR}_0^\delta}{\text{mr}^{1+\delta}}\ .....(\text{i})$
$\text{mvr}=\text{nh}$
$\Rightarrow\ \text{r}=\frac{\text{nh}}{\text{mv}}=\frac{\text{nh}}{\text{m}}\Big[\frac{\text{m}}{\text{xR}_0^\delta}\Big]^{\frac{1}{2}}\text{r}^{\frac{1+\delta}{2}}$
$[$Applying Bohr's second postulates$]$
Solving this for $r,$ we get $\text{r}_\text{n}=\Big[\frac{\text{n}^2\text{h}^2}{\text{mxR}_0^\delta}\Big]^\frac{1}{1-\delta}$
where $, rn$ is the radius of nth orbit of electron.
For $n = 1$ and substituting the values of constant, we get
$\text{r}_1=\Big[\frac{\text{n}^2\text{h}^2}{\text{mxR}_0^\delta}\Big]^\frac{1}{1-\delta}$
$\Rightarrow\ \text{r}_1=\Big[\frac{1.05^2\times10^{-68}}{9.1\times10^{-31}\times2.3\times10^{-28}\times10^{+19}}\Big]^\frac{1}{2.9}$
$=8\times10^{-11}=0.08\text{nm}(< 0.1\text{nm})$
This is the radius of orbit of electron in ground state of hydrogen atom.
Again using Bohr's second postulate, the speed of electron
$\text{v}_\text{n}=\frac{\text{nh}}{\text{mr}_\text{n}}=\text{nh}\Big(\frac{\text{mxR}_0^\delta}{\text{n}^2\text{h}^2}\Big)^\frac{1}{1-\delta}$
For $n = 1,$ the speed of electron in ground state $\text{v}_1=\frac{\text{h}}{\text{mr}_1}=1.44\times10^{6}\text{m/s}$
The kinetic energy of electron in ground state
$\text{KE}=\frac{1}{2}\text{mv}_1^2-9.43\times10^{-19}\text{J}=5.9\text{eV}$
Potential energy of electron in ground state till $R_0$
$\text{U}=\int\limits_0^{\text{R}_0}\text{Fdr}=\int\limits_0^{\text{R}_0}\text{dr}\approx-\frac{\text{x}}{\text{R}_0}$
Potential energy from $R_0$ to $r,$ $\text{U}=\int\limits_{\text{R}_0}^{\text{r}}\text{Fdr}=\int\limits_{\text{R}_0}^{\text{r}}\frac{\text{xR}_0^\delta}{\text{r}^{2+\delta}}\text{dr}^{1}$
$\text{U}=+\text{xR}_0^{\delta}\int^\text{r}_{\text{R}_0}\frac{\text{dr}}{\text{r}^{2+\delta}}=+\frac{\text{xR}_0^\delta}{-1-\delta}\Big[\frac{1}{\text{r}^{1+\delta}}\Big]^\text{r}_{\text{R}_0}$
$\text{U}=\frac{\text{xR}_0^\delta}{1+\delta}\Big[\frac{1}{\text{r}^{1+\delta}}-\frac{1}{\text{R}_0^{1+\delta}}\Big]=-\frac{\text{x}}{1+\delta}\Big[\frac{\text{R}^{\delta}}{\text{r}^{1+\delta}}-\frac{1}{\text{R}_0}\Big]$
$\text{U}=-\text{x}\Big[\frac{\text{R}_0^\delta}{\text{r}^{1+\delta}}-\frac{1}{\text{R}_0}+\frac{1+\delta}{\text{R}_0}\Big]$
$\text{U}=-\text{x}\Big[\frac{\text{R}_0^{-19}}{\text{r}^{-0.9}}-\frac{1.9}{\text{R}_0}\Big]$
$=\frac{2.3}{0.9}\times10^{-18}[(0.8)^{0.9}-1.9]\text{J}=-17.3\text{eV}$
Hence total energy of electron in ground state $= (-17.3 + 5.9) = -11.4eV.$
View full question & answer→Question 315 Marks
Calculate the smallest wavelength of radiation that may be emitted by $(a)$ hydrogen$, (b) \ He^+$ and $(c) \ Li^{++}$
AnswerSmall wave length is emitted i.e. longest energy, $\text{n}_1=1,\text{ n}_2=\infty$
- $\frac{1}{\lambda}=\text{R}\bigg(\frac{1}{\text{n}^2_1-\text{n}^2_2}\bigg)$
$\Rightarrow\frac{1}{\lambda}=1.1\times10^7\Big(\frac{1}{1}-\frac{1}{\infty}\Big)$
$\Rightarrow\lambda=\frac{1}{1.1\times10^7}=\frac{1}{1.1}\times10^{-7}$
$\Rightarrow\lambda=0.909\times10^{-7}$
$\Rightarrow\lambda=90.9\times10^{-8}=91\ \text{nm}$
- $\frac{1}{\lambda}=\text{z}^2\text{R}\bigg(\frac{1}{\text{n}^2_1}-\frac{1}{\text{n}^2_2}\bigg)$
$\lambda=\frac{1}{1.1\times10^{-7}\text{z}^2}=\frac{91\text{nm}}{4}=23\ \text{nm}$
- $\frac{1}{\lambda}=\text{z}^2\text{R}\bigg(\frac{1}{\text{n}^2_1}-\frac{1}{\text{n}^2_2}\bigg)$
$\Rightarrow\lambda=\frac{91\text{nm}}{\text{z}^2}=\frac{91}{9}=10\ \text{nm}$ View full question & answer→Question 325 Marks
A group of hydrogen atoms are prepared in n = 4 states. List the wavelength that are emitted as the atoms make transitions and return to n = 2 states.
Answer$\text{n}_1-4\xrightarrow{\ \ \ }\text{n}_2=2$
$\text{n}_1=4\xrightarrow{\ \ \ }3\xrightarrow{\ \ \ }2$
$\frac{1}{\lambda}=1.097\times10^7\Big(\frac{1}{16}-\frac{1}{4}\Big)$
$\Rightarrow\frac{1}{\lambda}=1.097\times10^7\Big(\frac{1-4}{16}\Big)$
$\Rightarrow\frac{1.097\times10^7\times3}{16}$
$\Rightarrow\lambda=\frac{16\times10^{-7}}{3\times1.097}=4.8617\times10^{-7}$
$\Rightarrow\lambda=1.861\times10^{-9}=487\text{nm}$
$\text{n}_1 =4\text{ and }\text{n}_2=3$
$\frac{1}{\lambda}=1.097\times10^7\Big(\frac{1}{16}-\frac{1}{9}\Big)$
$\Rightarrow\frac{1}{\lambda}=1.097\times10^7\Big(\frac{9-16}{144}\Big)$
$\Rightarrow\frac{1.097\times10^7\times7}{144}$
$\Rightarrow\lambda=\frac{144}{7\times1.097\times10^7}=1875\text{nm}$
$\text{n}_1=3\xrightarrow{\ \ \ }\text{n}_2=2$
$\frac{1}{\lambda}=1.097\times10^7\Big(\frac{1}{9}-\frac{1}{4}\Big)$
$\Rightarrow\frac{1}{\lambda}=1.097\times10^7\Big(\frac{4-9}{36}\Big)$
$\Rightarrow\frac{1.097\times10^7\times5}{66}$
$\Rightarrow\lambda=\frac{36\times10^{-7}}{5\times1.097}=656\text{nm}$
View full question & answer→Question 335 Marks
Radiation coming from transition $n = 2$ to $n = 1$ of hydrogen atoms falls on helium ions in $n = 1$ and $n = 2$ states. What are the possible transitions of helium ions as they absorbs energy from the radiation?
AnswerFrom transitions $n = 2$ to $n = 1$
$\text{E}=13.6\Big(\frac{1}{1}-\frac{1}{4}\Big)=13.6\times\frac{3}{4}=10.2\text{eV}$
Let in check the transitions possible on He.
$\text{n} = 1 \text{ to n} = 2$
$\text{E}_1=4\times13.6\Big(1-\frac{1}{4}\Big)=40.8\text{eV} [E_1 > E$ hence it is not possible$]$
$\text{n} = 1 \text{ to n} = 3$
$\text{E}_2=4\times13.6\Big(1-\frac{1}{9}\Big)=48.3\text{eV} [E_2 > E$ hence impossible$]$
Similarly $n = 1$ to $n = 4$ is also not possible.
$\text{n} = 2 \text{ to n} = 3$
$\text{E}_3=4\times13.6\Big(\frac{1}{4}-\frac{1}{9}\Big)=7.56\text{eV}$
$\text{n} = 2 \text{ to n} = 4$
$\text{E}_4=4\times13.6\Big(\frac{1}{4}-\frac{1}{16}\Big)10.2\text{eV}$
As, $E_3 < E$ and $E_4 = E$
Hence $E_3$ and $E_4$ can be possible.
View full question & answer→Question 345 Marks
A neutron moving with a speed $υ$ strikes a hydrogen atom in ground state moving towards it with the same speed. Find the minimum speed of the neutron for which inelastic $($completely or partially$)$ collision may take place. The mass of neutron $=$ mass of hydrogen $= 1.67 \times 10^{-27}kg$.
AnswerEnergy of the neutron is $\frac{1}{2}\text{mv}^2$
The condition for inelastic collision is,
$\frac{1}{2}\text{mv}^2>2\Delta\text{E}$
$\Delta\text{E}=\frac{1}{2}\text{mv}^2$
$\Delta\text{E}$ is the energy absorbed.
Energy required for first excited state is $10.2ev.$
$\therefore\ \Delta\text{E}<10.2\text{ev}$
$\therefore10.2\text{ev}<\frac{1}{2}\text{mv}^2$
$\text{V}_\text{min}=\sqrt{\frac{4\times10.2}{\text{m}}}\text{ev}$
$\therefore\text{v}=\sqrt{\frac{10.2\times1.6\times10^{-19}\times4}{1.67\times10^{-27}}}=6\times10^4\text{m/sec}$
View full question & answer→Question 355 Marks
The difference in the frequencies of series limit of Lyman series and Balmer series is equal to the frequency of the first line of the Lyman series. Explain.
AnswerThe 'series limit' refers to the 'shortest wavelength' $($corresponding to the maximum photon energy$)$.
The frequency of the radiation emitted for transition from $n_1 $ to $n_2$ is given by,
$\text{f}=\text{k}\bigg(\frac{1}{\text{n}^{2}_1}-\frac{1}{\text{n}^{2}_2}\bigg)$
Here $, k$ is a constant.
For the series limit of Lyman series,
$\text{n}_1=1$
$\text{n}_2=\infty$
Frequency, $\text{f}_1=\text{k}\Big(\frac{1}{1^2}-\frac{1}{\infty}\Big)=\text{k}$
For the first line of Lyman series,
$\text{n}_1=1$
$\text{n}_2=2$
Frequency, $\text{f}_2=\text{k}\Big(\frac{1}{1^2}-\frac{1}{2^2}\Big)=\frac{3\text{k}}{4}$
For series limit of Balmer series,
$\text{n}_1=2$
$\text{n}_2=\infty$
$\text{f}_1=\text{k}\Big(\frac{1}{2^2}-\frac{1}{\infty}\Big)=\frac{\text{k}}{4}$
$\text{f}_1-\text{f}_3=\text{f}_2$
Thus, the difference in the frequencies of series limit of Lyman series and Balmer series is equal to the frequency of the first line of the Lyman series.
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