5 Marks Questions - Physics STD 12 Science Questions - Vidyadip

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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.

Answer

Radius 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 × 10^-34 Js
me = Mass of an electron = 9.1 × 10^-31kg
e = Charge of an electron = 1.9 × 10^-19 C
mp = Mass of a proton = 1.67 × 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 × 10^-11 N m²/ kg²
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 × 10²⁷ 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.

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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?

Answer

It 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 10⁷ 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.

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Question 35 Marks

Obtain an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n – ncy of revolution of the electron in the orbit.

Answer

It 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,
v₁ = 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 (E₂) 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,
v₂ = Frequency of radiation at level (n - 1)
Energy (E) released as a result of de-excitation:
E = E₂ -E₁
hv = E₂ -E₁...(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 $(2\text{n}-1)\approx2\text{n}\text{ and }(\text{n}-1)\approx\text{n}$
$\therefore\ \text{v}=\frac{\text{me}^4}{32\pi\in^2_0\Big(\frac{\text{h}}{2\pi}\Big)^3\text{n}^3}\ \dots(\text{iv)}$
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.

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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.

Answer

For ground level, n₁ = 1
Let E₁ be the energy of this level. It ls known that E₁ Is related with n₁ 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, n₂ = 4.
Let E₂ 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 = E₂ - E₁
$=\frac{-13.6}{16}-\Big(-\frac{13.6}{1}\Big)$
$=\frac{13.6\times15}{16}\text{ eV}$
$=\frac{13.6\times15}{16}\times1.6\times10^{-19}=2.04\times10^{-18}\text{ J}$
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 10⁸m/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 10¹⁵ Hz.

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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 v₁ be the orbital speed of the electron in a hydrogen atom in the ground state level, n₁ = 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 × 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 × 10⁸ = 2.18 × 10⁶ m/s

For level n₂= 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 × 10⁶ m/s

And, for n₃ = 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 × 10⁵ m/s

Hence, the speed of the electron in a hydrogen atom in n = 1, n = 2 and n = 3 is 2.18 × 10⁶ m/s, 1.09 × 10⁶ m/s, 7.27 × 10⁵ m/s respectively.

Let T₁ be the orbital period of the electron when it is in level n₁ = 1. Orbital period is related to orbital speed as:

$\text{T}_1=\frac{2\pi\text{r}_1}{\text{v}_1}$

Where,

r₁= 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 C²m^-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 × 10^-17 = 1.527 × 10^-16 s

For level n₂ = 2, we can write the period as:

$\therefore\ \text{T}_2=\frac{2\pi\text{r}_2}{\text{v}_2}$

Where,

r₂ = Radius of the electron in n₂ = 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 × 10^-15 s

And, for level n₃ = 3, we can write the period as:

$\therefore\ \text{T}_3=\frac{2\pi\text{r}_3}{\text{v}_3}$

Where,

r₃= Radius of the electron in n₃ = 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 × 10^-15 s

Hence, the orbital period in each of these levels is 1.52 x 10^-16 s, 1.22 x 10^-15s, and 4.12 × 10-¹⁵ s respectively.

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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^–10m).

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 × 10^-19 C

Mass of an electron, me = 9.1 × 10^-31 kg

Speed of light, c = 3 × 10⁸ 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 × 10^-19C

Mass of an electron, me = 9.1 × 10^-31 kg

Planck’s constant, h = 6.63 × 10^-34Js

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.

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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?

Answer

The 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.

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Question 85 Marks

Calculate the smallest wavelength of radiation that may be emitted by (a) hydrogen, (b) He⁺ and (c) Li⁺⁺

Answer

Small wave length is emitted i.e. longest energy,

$\text{n}_1=1,\text{ n}_2=\infty$

$\frac{1}{\lambda}=\text{R}\bigg(\frac{1}{\text{n}^2_1-\text{n}^2_2}\bigg)$

$\Rightarrow\frac{1}{\lambda}=1.1\times10^7\Big(\frac{1}{1}-\frac{1}{\infty}\Big)$

$\Rightarrow\lambda=\frac{1}{1.1\times10^7}=\frac{1}{1.1}\times10^{-7}$

$\Rightarrow\lambda=0.909\times10^{-7}$

$\Rightarrow\lambda=90.9\times10^{-8}=91\text{nm}$

$\frac{1}{\lambda}=\text{z}^2\text{R}\bigg(\frac{1}{\text{n}^2_1}-\frac{1}{\text{n}^2_2}\bigg)$

$\lambda=\frac{1}{1.1\times10^{-7}\text{z}^2}=\frac{91\text{nm}}{4}=23\text{nm}$

$\frac{1}{\lambda}=\text{z}^2\text{R}\bigg(\frac{1}{\text{n}^2_1}-\frac{1}{\text{n}^2_2}\bigg)$

$\Rightarrow\lambda=\frac{91\text{nm}}{\text{z}^2}=\frac{91}{9}=10\text{nm}$

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Question 95 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

$\text{n}_1-4\xrightarrow{\ \ \ }\text{n}_2=2$
$\text{n}_1=4\xrightarrow{\ \ \ }3\xrightarrow{\ \ \ }2$
$\frac{1}{\lambda}=1.097\times10^7\Big(\frac{1}{16}-\frac{1}{4}\Big)$
$\Rightarrow\frac{1}{\lambda}=1.097\times10^7\Big(\frac{1-4}{16}\Big)$
$\Rightarrow\frac{1.097\times10^7\times3}{16}$
$\Rightarrow\lambda=\frac{16\times10^{-7}}{3\times1.097}=4.8617\times10^{-7}$
$\Rightarrow\lambda=1.861\times10^{-9}=487\text{nm}$
$\text{n}_1 =4\text{ and }\text{n}_2=3$
$\frac{1}{\lambda}=1.097\times10^7\Big(\frac{1}{16}-\frac{1}{9}\Big)$
$\Rightarrow\frac{1}{\lambda}=1.097\times10^7\Big(\frac{9-16}{144}\Big)$
$\Rightarrow\frac{1.097\times10^7\times7}{144}$
$\Rightarrow\lambda=\frac{144}{7\times1.097\times10^7}=1875\text{nm}$
$\text{n}_1=3\xrightarrow{\ \ \ }\text{n}_2=2$
$\frac{1}{\lambda}=1.097\times10^7\Big(\frac{1}{9}-\frac{1}{4}\Big)$
$\Rightarrow\frac{1}{\lambda}=1.097\times10^7\Big(\frac{4-9}{36}\Big)$
$\Rightarrow\frac{1.097\times10^7\times5}{66}$
$\Rightarrow\lambda=\frac{36\times10^{-7}}{5\times1.097}=656\text{nm}$

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Question 105 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?

Answer

From 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.
$\text{n} = 1 \text{ to n} = 2$
$\text{E}_1=4\times13.6\Big(1-\frac{1}{4}\Big)=40.8\text{eV}$ [E₁ > E hence it is not possible]
$\text{n} = 1 \text{ to n} = 3$
$\text{E}_2=4\times13.6\Big(1-\frac{1}{9}\Big)=48.3\text{eV}$ [E₂ > 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, E₃ < E and E₄ = E
Hence E₃ and E₄ can be possible.

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Question 115 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 × 10^-27kg.

Answer

Energy 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.
$\therefore\ \Delta\text{E}<10.2\text{ev}$
$\therefore10.2\text{ev}<\frac{1}{2}\text{mv}^2$
$\text{V}_\text{min}=\sqrt{\frac{4\times10.2}{\text{m}}}\text{ev}$
$\therefore\text{v}=\sqrt{\frac{10.2\times1.6\times10^{-19}\times4}{1.67\times10^{-27}}}=6\times10^4\text{m/sec}$

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Question 125 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.

Answer

The 'series limit' refers to the 'shortest wavelength' (corresponding to the maximum photon energy).
The frequency of the radiation emitted for transition from n₁ to n₂is 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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Question 135 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 ν₀ emitted if the atom were at rest.

Answer

Velocity 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
$1-\frac{\text{u}}{\text{c}}=\text{q}$
$\therefore\ \text{v}=\text{v}_0\bigg(\frac{1+\frac{\text{u}}{\text{c}}}{1}\bigg)=\text{v}_0\Big(1+\frac{\text{u}}{\text{c}}\Big)$
$\text{v}=\text{v}_0\Big(1+\frac{\text{u}}{\text{c}}\Big)$

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Question 145 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,

$\frac{\text{hc}}{\lambda}-1.9=\frac{13.6\times5}{36}$

$\frac{\text{hc}}{\lambda}=\frac{13.6\times5}{36}+1.9$

$\frac{1240}{\lambda1}=1.88+1.9=3.78$

$\lambda=\frac{1240}{3.78}$

$\lambda=328.04\text{nm}$

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Question 155 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.

Answer

Key concept: Total energy (for Hydrogen and H₂ 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,
$\text{E}=-\Big(\frac{\text{me}^4}{8\in_0\text{h}^2}\Big).\frac{\text{Z}^2}{\text{n}^2}=-\Big(\frac{\text{me}^2}{8\in_0\text{ch}^3}\Big)\text{ch}\frac{\text{Z}^2}{\text{n}^2}$
$=-\text{R ch}\frac{\text{Z}^2}{\text{n}^2}=-13.6\frac{\text{Z}^2}{\text{n}^2}\text{eV}$
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 × 13.6)eV = 54.5eV.

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Question 165 Marks

The inverse square law in electrostatics is $|\text{F}|=\frac{\text{e}^2}{(4\pi\in_0)\text{r}^2}$ 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 $|\text{F}|=\frac{\text{e}^2}{(4\pi\in_0)\text{r}^2}\Big[\frac{1}{\text{r}^2}+\frac{\lambda}{\text{r}}\Big].\text{e}\times\text{p}(-\lambda\text{r})$ 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.

Answer

We 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 $|\text{F}|=\Big(\frac{\text{e}^2}{4\pi\in_0}\Big)\bigg[\frac{1}{\text{r}^2}+\frac{\lambda}{\text{r}}\bigg]\text{e}^{\lambda\text{r}}\ .....(\text{i})$
and $|\text{F}|=\frac{\text{dU}}{\text{dr}},$
$\text{U}_\text{r}=\int|\text{F}|\text{dr}=\Big(\frac{\text{e}^2}{4\pi\in_0}\Big)\int\Big(\frac{\lambda_\text{e}^{-\lambda\text{r}}}{\text{r}}+\frac{\text{e}^{-\lambda\text{r}}}{\text{r}^2}\Big)\text{dr}$
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
$\text{mvr}=\text{h}\Rightarrow\ \text{v}=\frac{\text{h}}{\text{mr}};\text{ and }$
$\frac{\text{mv}^2}{\text{r}}=\text{F}=\Big(\frac{\text{e}^2}{4\pi\in_0}\Big)\Big(\frac{1}{\text{r}^2}+\frac{\lambda}{\text{r}}\Big)$
[Putting $\text{e}^{-\lambda\text{r}}\approx1$ in eqn. (i)]
Thus, $\Big(\frac{\text{m}}{\text{r}}\Big)\Big(\frac{\text{h}^2}{\text{m}^2\text{r}^2}\Big)=\Big(\frac{\text{e}^2}{4\pi\in_0}\Big)\Big(\frac{1}{\text{r}^2}+\frac{\lambda}{\text{r}}\Big)$
or $\frac{\text{h}^2}{\text{m}}=\Big(\frac{\text{e}^2}{4\pi\in_0}\Big)\Big(\frac{\text{r}+\lambda\text{r}^2}{\text{r}^3}\Big)$
or $\frac{\text{h}^2}{\text{m}}=\Big(\frac{\text{e}^2}{4\pi\in_0}\Big)({\text{r}+\lambda\text{r}^2})\ .....(\text{iii})$
When $\lambda=0,\text{r}=\text{r}'_\text{B}$ and
$\frac{\text{h}^2}{\text{m}}=\Big(\frac{\text{e}^2}{4\pi\in_0}\Big)\text{r}_\text{B}\ .....(\text{iv})$
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) $\text{u}_\text{r}=-\Big(\frac{\text{e}^2}{4\pi\in_0}\Big)\frac{\text{e}^{-\lambda(\text{r}_\text{B}+\delta)}}{(\text{r}_\text{B}+\delta)}$
$=-\Big(\frac{\text{e}^2}{4\pi\in_0}\frac{1}{\text{r}_\text{B}}\Big)\Big(1-\frac{\delta}{\text{r}_\text{B}}\Big)(1-\lambda\text{r}_\text{B})\approx-\frac{\text{e}^2}{4\pi\in_0\text{r}_\text{B}}$
$=-24.2\text{eV}$
$[\because\ \text{e}^{-\lambda(\text{r}_\text{B}+\delta)}\approx\ 1-\lambda(\text{r}_\text{B}+\delta)=1-\lambda\text{r}_\text{B}-\lambda\delta\approx1-\lambda\text{e}_\text{B}]$
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
$=(-13.6+27.2\lambda\text{r}_\text{B})\text{eV}-(-13.6\text{eV})$
$=(27.2\lambda\text{r}_\text{B})\text{eV}$

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Question 175 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 = E₁ - E₅ = 13.6 - 0.54 = 13.06eV
If angular momentum is conserved, angular momentun of photon = change in angular momentum of electron = L₅ - L₂ = 5h - 2h = 3 × 1.06 × 10^-34 = 3.18 × 10^-34 = 3.18 × 10^-38kh m²/s.

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Question 185 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 $\text{E}_\text{n}=\frac{\text{E}_1}{\text{n}^2}?\text{ r}_\text{n}=\text{a}_0\text{n}^2?$

Answer

Energy 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.

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Question 195 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 × 10^-24kg. Mass of the sun = 2.0 × 10³⁰kg, earth-sun distance = 1.5 × 10¹¹m.

Answer

Mass of Earth = Me = 6.0 × 1024kg

Mass of Sun = Ms = 2.0 × 1030kg

Earth-Sun dist = 1.5 × 1011m

$\text{mvr}=\frac{\text{nh}}{2\pi}\text{ or m}^2\text{v}^2\text{r}^2=\frac{\text{n}^2\text{h}^2}{4\pi^2}\ ....(\text{i})$

$\frac{\text{GMeMs}}{\text{r}^2}=\frac{\text{mev}^2}{\text{r}}\text{ or v}^2=\frac{\text{GM}}{\text{s/r}}\ ....(\text{ii})$

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

$\text{r}=\sqrt{\frac{\text{h}^2}{4\pi^2\text{GMsMe}^2}}=2.29\times10^{-138}\text{m}$

$\text{r}=2.3\times10^{-138}\text{m}$

$\text{n}^2=\frac{\text{Me}^2\times\text{r}\times4\times\pi^2\times\text{G}\times\text{Ms}}{\text{h}^2}$

$\text{n}^2=2.5\times10^{74}$

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Question 205 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?

Answer

When 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.

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Question 215 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?

Answer

Key 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
$\text{E}_\text{n}=-\left\{\frac{\mu}{2\text{h}^2}\Big(\frac{\text{e}^2}{4\pi\in_0}\Big)^2\right\}\frac{1}{\text{n}^2}$
$\mu$ 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)
$\text{E}_\text{n}=-\left\{\frac{\mu}{2\text{h}^2}\Big(\frac{\text{e}^2}{4\pi\in_0}\Big)^2\right\}\frac{1}{\text{n}^2}=-\frac{13.6\text{eV}}{\text{n}^2}=\frac{-6.8\text{eV}}{\text{n}^2}$
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.

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Question 225 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 = 0.1Å.
R = 10 Å.

Answer

For a point nucleus in H-atom,
Ground state: mvr = h,
$\frac{\text{mv}^2}{\text{r}_\text{B}}=-\frac{\text{e}^2}{\text{r}^2_\text{B}}.\frac{1}{4\pi\in_0}$
The electrostatic force of attraction between positively charged nucleus and negatively charged electrons (Coulombian force) provides necessary centripetal force of revolution.
$\frac{\text{mv}^2}{\text{r}_\text{B}}=-\frac{\text{e}_2}{\text{r}^2_\text{B}}.\frac{1}{4\pi\in_0}$
By Bohr's postulates in ground state, we have
mvr = h
$\therefore\ \text{m}\frac{\text{h}^2}{\text{m}^2\text{r}^2_\text{B}}=+\Big(\frac{\text{e}^2}{4\pi\in_0}\Big)\frac{1}{\text{r}_\text{B}^2}$
$\therefore\ \frac{\text{h}^2}{\text{m}}.\frac{4\pi\in_0}{\text{e}^2}=\text{r}_\text{B}=0.51\mathring{\text{A}}$
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, $\text{r}_\text{B}^4=\text{e}\Big(\frac{\text{r}^3_\text{B}}{\text{R}^3}\Big)$
$\therefore\ \text{r}'_\text{B}=\frac{\text{h}^2}{\text{m}}\Big(\frac{4\pi\in_0}{\text{e}^2}\Big)\frac{\text{R}^3}{\text{r}_\text{B}'^3}$
$\text{r}_\text{B}'^4=(0.51\mathring{\text{A}}),\text{R}^3\ \ [\text{R}=10\mathring{\text{A}}]$
$=510(\mathring{\text{A}})^4$
$\therefore\ \text{r}'\text{B}\approx(510)^{\frac{1}{4}}\mathring{\text{A}}<\text{R}$
$\text{KE}=\frac{1}{2}\text{mv}^2=\frac{\text{m}}{2}.\frac{\text{h}}{\text{m}^2\text{r}'^2_\text{B}}=\frac{\text{h}}{2\text{m}}.\frac{1}{\text{r}'^2_\text{B}}$
$=\Big(\frac{\text{h}}{2\text{mr}^2_\text{B}}\Big)\Big(\frac{\text{r}^2_\text{B}}{\text{r}'^2_\text{B}}\Big)=(13.6\text{eV})\frac{(0.51)^2}{(510)^\frac{1}{2}}=\frac{3.54}{22.6}=0.16\text{eV}$
$\text{PE}=+\Big(\frac{\text{e}^2}{4\pi\in_0}\Big).\Big(\frac{\text{r}'^2-3\text{R}^2}{2\text{R}^3}\Big)=+\Big(\frac{\text{e}^2}{4\pi\in_0}.\frac{1}{\text{r}_\text{B}}\Big)\bigg(\frac{\text{r}_\text{B}(\text{r}'^2_\text{B}-3\text{R}^2)}{\text{R}^3}\bigg)$
$=+(27.2\text{eV})\bigg[\frac{0.51(\sqrt{510}-300)}{1000}\bigg]$
$=+(27.2\text{eV}),\frac{-141}{1000}=-3.83\text{eV}$

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Question 235 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 μ, 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 × 10^-27kg, Mass of ²H nucleus is 3.3374 × 10^-27kg, Mass of electron = 9.109 × 10^-31kg).

Answer

According 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

$\text{E}_\text{n}=-\left\{\frac{\mu}{2\text{h}^2}\Big(\frac{\text{e}^2}{4\pi\in_0}\Big)^2\right\}\frac{\text{Z}^2}{\text{n}^2}=-\frac{\mu\text{Z}^2\text{e}^4}{8\in_0^2\text{h}^2}\Big(\frac{1}{\text{h}^2}\Big)$

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 $\text{hf}_\text{H}=\frac{\mu\text{Z}^2\text{e}^4}{8\in_0^2\text{h}^2}\Big(1-\frac{1}{4}\Big)=\frac{\mu_\text{H}\text{e}^4}{8\in_0^2\text{h}^2}\times\frac{3}{4}$

Thus, the wavelength of the transition is $\lambda\text{H}=\frac{3}{4}\frac{\mu_\text{H}\text{e}^4}{8\in_0^2\text{h}^3\text{c}}$. The wavelength of the treansition for the same line in Deutrium is $\lambda_\text{D}=\frac{3}{4}\frac{\mu_\text{D}\text{e}^4}{8\in_0^2\text{h}^3\text{c}}$.

$\therefore$ 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.

$\Rightarrow\ \frac{\Delta\lambda}{\lambda_\text{H}}\times100\simeq\Big[\frac{\text{m}_\text{e}}{\text{M}_\text{H}}-\frac{\text{m}_\text{e}}{\text{M}_\text{D}}\Big]\times100$

$\Rightarrow\ \frac{\Delta\lambda}{\lambda_\text{H}}\times100\approx\Big[\frac{1}{\text{M}_\text{H}}-\frac{1}{\text{M}_\text{D}}\Big]\times100$

$=9.1\times10^{-31}\bigg[\frac{1}{1.6725\times10^{-27}}-\frac{1}{3.3374\times10^{-27}}\bigg]\times100$

$=9.1\times10^{-4}[0.5979-0.2996]\times100=2.714\times10^{-2}\%$

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Question 245 Marks

How many wavelengths are emitted by atomic hydrogen in visible range (380nm-780nm)? In the range 50nm to 100nm?

Answer

Balmer series contains wavelengths ranging from 364nm (for n₂ = 3) to $655\text{nm }(\text{n}_2)=\infty$ So, the given range of wavelength (380-780nm) 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 × 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 91nm (for n₂ = 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 ton = 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 50nm to 100nm are 3

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Question 255 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 E₁ to E₂

E₁ to E₂, energy absorbed = 10.2eV

Energy left = 12.42 - 10.2 = 2.22eV

$2.22\text{eV}=\frac{\text{hc}}{\lambda}=\frac{1242}{\lambda}$ or $\lambda=559.45=560\text{nm}$

E₁ to E₃, Energy absorbed = 12.1eV

Energy left = 12.42 - 12.1 = 0.32eV

$0.32=\frac{\text{hc}}{\lambda}=\frac{1242}{\lambda}$ or $\lambda=\frac{1242}{0.32}=3881.2=3881\text{nm}$

E₃ to E₄, 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.

$10.2=\frac{\text{hc}}{\lambda}\text{ or }\lambda=\frac{1242}{10.2}=121.76\text{nm}$

$12.1=\frac{\text{hc}}{\lambda}\text{ or }\lambda=\frac{1242}{12.1}=102.64\text{nm}$

$0.65=\frac{\text{hc}}{\lambda}\text{ or }\lambda=\frac{1242}{0.65}=1910.76\text{nm}$

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Question 265 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 $=\text{KE}+\text{PE}=\frac{2\pi^2\text{G}^2\text{M}^2_\text{n}\text{M}^3_\text{e}}{2\text{n}^2\text{h}^2}$

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Question 275 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.

Answer

From 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 × 10⁷m^-1

n₁ = Quantum number of final state

n₂ = Quantum number of initial state

For transition from n = 3 to n = 2,

Here, n₁ = 2, n₂ = 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, n₁ = 4, n₂ = 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, n₁ = 9, n₂ = 10

$\frac{1}{\lambda}=1.09737\times10^7\Big(\frac{1}{81}-\frac{1}{100}\Big)$

$\lambda=\frac{81\ \times\ 100}{19\ \times\ 1.09737\ \times\ 10^7}$

$\lambda=38849\text{nm}$

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Question 285 Marks

The first four spectral lines in the Lyman serics of a H-atom are λ = 1218Å, 1028Å, 974.3Å and 951.4Å. If instead of Hydrogen, we consider Deuterium, calculate the shift in the wavelength of these lines.

Answer

Let µ_H and µ_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.
$\lambda\propto\frac{1}{\text{r}}\text{ or }\frac{\lambda_\text{D}}{\lambda_\text{H}}=\frac{\text{R}_\text{H}}{\text{R}_\text{D}}\ .....(\text{i})$
$\text{R}_\text{R}=\frac{\text{m}_\text{e}\text{e}^4}{8\in_0\text{ch}^3}=\frac{\mu_\text{H}\text{e}^4}{8\in_0\text{ch}^3}$
$\text{R}_\text{H}=\frac{\text{m}_\text{e}\text{e}^4}{8\in_0\text{ch}^3}=\frac{\mu_\text{H}\text{e}^4}{8\in_0\text{ch}^3}$
$\therefore\ \frac{\text{R}_\text{H}}{\text{R}_\text{D}}=\frac{\text{R}_\text{H}}{\text{R}_\text{D}}\ .....(\text{ii})$
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 $(\lambda_\text{H}-\lambda_\text{D})\approx0.3\mathring{\text{A}}.$

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Question 295 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.

Answer

According 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]

$\text{v}^{2}=\frac{\text{heB}}{2\pi\text{m}^2}$

$\text{v}=\sqrt{\frac{\text{heB}}{2\pi\text{m}^2}}$

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Question 305 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 × 10^-27kg.

Answer

The hydrogen atoms after collision move with speeds v₁ and v₂
$\text{mv}=\text{mv}_1+\text{mv}_2\ ....(\text{i})$
$\frac{1}{2}\text{mv}^2=\frac{1}{2}\text{mv}^2_1+\frac{1}{2}\text{mv}^2_2+\Delta\text{E}\ ....(\text{ii})$
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’
$\text{v}_1=\text{v}_2$
$\text{v}^2-\Big(\frac{4\Delta\text{E}}{\text{m}}\Big)=0$
$\text{v}^2=\frac{4\Delta\text{E}}{\text{m}}=\frac{4\times13.6\times1.6\times10^{-19}}{1.67\times10^{-27}}$
$\text{v}=\sqrt{\frac{4\times13.6\times1.6\times10^{-19}}{1.67\times10^{-27}}}=7.2\times10^4\text{m/s}$

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Question 315 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 +q₁, -q₂ is modified to
$|\text{F}|=\frac{\text{q}_1\text{q}_2}{(4\pi\in_0)}\frac{1}{\text{r}^2}.\text{r}\geq\text{R}_0$
$=\frac{\text{q}_1\text{q}_2}{(4\pi\in_0)}\frac{1}{\text{R}^2_0}\Big(\frac{\text{R}_0}{\text{r}}\Big)^{\in}.\text{r}\geq\text{R}_0$
Calculate in such a case, the ground state energy of a H-atom, if $\in=0.1,\text{R}_0=1\mathring{\text{A}}$.

Answer

Leu us consider the case, when $\text{r}\leq\text{R}_0=1.\mathring{\text{A}}$
Let $\in=2+\delta$
$\text{F}=\frac{\text{q}_1\text{q}_2}{4\pi\in_0}.\frac{\text{R}_\delta}{\text{r}^{2+\delta}}=\frac{\text{xR}_0^\delta}{\text{r}^{2+\delta}}$
Where, $\frac{\text{q}_1\text{q}_2}{4\pi_0\in_0}=\text{x}=(1.6\times10^{-19})^2\times9\times10^9$
$=2.04\times10^{-29}\text{Nm}^2$
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 R₀
$\text{U}=\int\limits_0^{\text{R}_0}\text{Fdr}=\int\limits_0^{\text{R}_0}\text{dr}\approx-\frac{\text{x}}{\text{R}_0}$
Potential energy from R₀ 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.

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Question 325 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.

Answer

As 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 $\text{KE}=\text{Z}^2\text{R}\Big(\frac{3}{4}-\frac{1}{16}\Big)=\frac{1}{16}\text{Z}^2\text{R}$

$=\frac{11}{16}\times24\times24\times13.6\text{eV}=5385.6\text{eV}.$

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Question 335 Marks

We have stimulated emission and spontaneous emission. Do we also have stimulated absorption and spontaneous absorption?

Answer

When a photon of energy (E₂ - E₁ = hυ) is incident on an atom in the ground state, the atom in the ground state E₁ may absorb the photon and jump to a higher energy state (E₂). 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.

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Question 345 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.

Answer

Key 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 n₂ to n₁ 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. n₂ = n and n₁ = 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.

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Question 355 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.

Answer

To 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

$\text{B}=\frac{\pi_0\text{e.e}^2\pi^2\text{m}^2\text{e}^4}{2\varepsilon_0\text{h}\times4\pi\times\varepsilon^2_0\text{h}^4}\Rightarrow\text{B}=\frac{\mu_0\text{e}^7\pi\text{m}^2}{8\varepsilon^3_0\text{h}^5}$

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