3 Marks Question - Physics STD 11 Science Questions - Vidyadip

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3 Marks Question

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19 questions · self-marked practice — reveal the answer and mark yourself.

Question 13 Marks

Find the temperature at which the average thermal kinetic energy is equal to the energy needed to take a hydrogen atom from its ground state to n = 3 state. Hydrogen can now emit red light of wavelength 653.1nm. Because of Maxwellian distribution of speeds, a hydrogen sample emits red light at temperatures much lower than that obtained from this problem. Assume that hydrogen molecules dissociate into atoms.

Answer

$\text{K}=8.62\times10^{-5}\text{eV}/\text{k}$K.E. of $H_2$ molecules $=\frac{3}{2}\text{KT}$
Energy released, when atom goes from ground state to no = 3
$\Rightarrow13.6\Big(\frac{1}{1}-\frac{1}{9}\Big)$
$\Rightarrow\frac{3}{2}\text{KT}=13.6\Big(\frac{1}{1}-\frac{1}{9}\Big)$
$\Rightarrow\frac{3}{2\ \times\ 8.62\ \times\ 10^{-5}\text{T}}=\frac{13.6\times8}{9}$
$\Rightarrow\text{T}=0.9349\times10^5=9.349\times10^4=9.4\times10^4\text{K}$

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Question 23 Marks

A hydrogen atom emits ultraviolet radiation of wavelength 102.5nm. What are the quantum numbers of the states involved in the transition?

Answer

As the light emitted lies in ultraviolet range the line lies in hyman series.$\frac{1}{\lambda}=\text{R}\bigg(\frac{1}{\text{n}^2_1}-\frac{1}{\text{n}^2_2}\bigg)$
$\Rightarrow\frac{1}{102.5\times10^{-9}}=1.1\times10^{7}\bigg(\frac{1}{1^2}-\frac{1}{\text{n}^2_2}\bigg)$
$\Rightarrow\frac{10^9}{102.5}=1.1\times10^7\bigg(1-\frac{1}{\text{n}^2_2}\bigg)$
$\Rightarrow\frac{10^2}{102.5}=1.1\times10^7\bigg(1-\frac{1}{\text{n}^2_2}\bigg)$
$\Rightarrow1-\frac{1}{\text{n}^2_2}=\frac{100}{102.5\times1.1}$
$\Rightarrow\frac{1}{\text{n}^2_2}=\frac{1-100}{102.5\times1.1}$
$\Rightarrow\text{n}_2=2.97=3$

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Question 33 Marks

A hydrogen atom in state $n = 6$ makes two successive transitions and reaches the ground state. In the first transition a photon of $1.13eV$ is emitted. $(a)$ Find the energy of the photon emitted in the second transition $(b)$ What is the value of n in the intermediate state?

Answer

Energy at $\text{n}=6,\ \text{E}=\frac{-13.6}{36}=-0.3777777$

Energy in groundstate $= -13.6eV$
Energy emitted in Second transition
$= -13.6 - (0.37777 + 1.13)$
$= -12.09$
$= 12.1eV$

Energy in the intermediate state $= 1.13eV + 0.0377777$

$=1.507777$
$=\frac{13.6\times\text{z}^2}{\text{n}^2}$
$=\frac{13.6}{\text{n}^2}$
or, $\text{n}=\sqrt{\frac{13.6}{1.507}}=3.03=3=\text{n}$

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Question 43 Marks

Which wavelengths will be emitted by a sample of atomic hydrogen gas (in ground state) if electrons of energy 12.2eV collide with the atoms of the gas?

Answer

As the electron collides, it transfers all its energy to the hydrogen atom. The excitation energy to raise the electron from the ground state to the nth state is given by,$\text{E}=\big(13.6\text{eV}\big)\times\Big(\frac{1}{1^2}-\frac{1}{\text{n}}\Big)$
Substituting n = 2, we get E = 10.2eV Substituting n = 3, we get E' = 12.08eV Thus, the atom will be raised to the second excited energy level. So, when it comes to the ground state, there is transitions from n = 3 to n = 1. Therefore, the wavelengths emitted will lie in the Lyman series (infrared region).

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Question 53 Marks

The first excited energy of a $He^+$ ion is the same as the ground state energy of hydrogen. Is it always true that one of the energies of any hydrogen-like ion will be the same as the ground state energy of a hydrogen atom?

Answer

The energy of hydrogen ion is given by,$\text{E}_\text{n}=\frac{136\text{eV}\text{Z}^2}{\text{n}^2}$
For the first excited state $(n = 2)$, the energy of $He^+$ ion (with Z = 2) will be $-13.6eV$. This is same as the ground state energy of a hydrogen atom. Similarly, for all the hydrogen like ions, the energy of the $(n - 1)^{th}$ excited state will be same as the ground state energy of a hydrogen atom if Z = n.

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Question 63 Marks

Average lifetime of a hydrogen atom excited to $n = 2$ state is $10^{-8}s$. Find the number of revolutions made by the electron on the average before it jumps to the ground state.

Answer

$\text{n}=2,\text{ T}=10^{-8}\text{s}$Frequancy $=\frac{\text{me}^4}{4\in^2_0\text{n}^3\text{h}^3}$
So, time period $=\frac{1}{\text{f}}=\frac{4\in_0^2\text{n}^3\text{h}^3}{\text{me}^4}$
$\frac{4\times(8.85)^2\times2^3\times(6.63)^3}{9.1\times(1.6)^4}\times\frac{10^{-24}-10^{-102}}{10^{-76}}$
$=12274.735\times10^{-19}\sec$
No.of revolutions $=\frac{10^{-8}}{12247.735\ \times\ 10^{-19}}=8.16\times10^5$
$=8.2\times10^6$ revolution.

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Question 73 Marks

Show that the ratio of the magnetic dipole moment to the angular momentum (l = mvr) is a universal constant for hydrogen-like atoms and ions. Find its value.

Answer

Magnetic Dipole moment $=\text{niA}=\frac{\text{e}\times\text{me}^4\times\pi\text{r}^2_\text{n}\text{n}^2}{4\in_0^2\text{h}^3\text{n}^3}$ Angular momentum $=\text{mvr}=\frac{\text{nh}}{2\pi}$ Since the ratio of magnetic dipole moment and angular momentum is independent of Z. Hence it is an universal constant. Ratio $=\frac{\text{e}^5\ \times\ \text{m}\ \times\ \pi\text{r}^2_0\text{n}^2}{24\in_0\text{h}^3\text{n}^3}\times\frac{2\pi}{\text{nh}}$$\Rightarrow\frac{\big(1.6\ \times\ 10^{-19}\big)^5\times\big(9.1\times10^{-31}\big)\times(3.14)^2\times\big(0.53\times10^{-10}\big)^2}{2\times\big(8.85\times10^{-12}\big)^2\times\big(6.63\times10^{-34}\big)^4\times1^2}$
$=8.73\times10^{10}\text{C/kg}$

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Question 83 Marks

When a photon is emitted from an atom, the atom recoils. The kinetic energy of recoil and the energy of the photon come from the difference in energies between the states involved in the transition. Suppose, a hydrogen atom changes its state from $n = 3$ to $n = 2$. Calculate the fractional change in the wavelength of light emitted, due to the recoil.

Answer

Difference in energy in the transition from n = 3 to n = 2 is 1.89eV (= E). If all this energy is used up in emitting a photon (i.e. recoil energy is zero). Then, $\text{E}=\frac{\text{hc}}{\lambda}$$\lambda=\frac{\text{hc}}{\text{E}}\ ...(\text{i})$
If difference of energy is used up in emitting a photon and recoil of atom, then let $E_R$ be the recoil energy of atom.$\text{E}=\frac{\text{hc}}{\lambda'}+\text{E}_\text{R}$
$\lambda'=\frac{\text{hc}}{\text{E}-\text{E}_\text{R}}\ ...(\text{ii})$
Fractional change in the wavelength is given as,$\frac{\Delta\lambda}{\lambda}=\frac{\lambda'-\lambda}{\lambda}$
$\frac{\Delta\lambda}{\lambda}=\frac{1}{\lambda}\Big(\frac{\text{hc}}{\text{E}-\text{E}_\text{R}}-\frac{\text{hc}}{\text{E}}\Big)$
$\frac{\Delta\lambda}{\lambda}=\frac{\text{E}}{\text{hc}}\Big(\frac{\text{hcE}_\text{R}}{\text{E}(\text{E}-\text{E}_\text{R})}\Big)$ $\Big(\because\ \lambda=\frac{\text{hc}}{\text{E}}\Big)$
$\frac{\Delta\lambda}{\lambda}=\Big(\frac{\text{E}_\text{R}}{\text{E}-\text{E}_\text{R}}\Big)$

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Question 93 Marks

Calculate the magnetic dipole moment corresponding to the motion of the electron in the ground state of a hydrogen atom.

Answer

Dipole moment $(\mu)$$=\text{niA}=\frac{1\times\text{q}}{\text{tA}}=\text{qfA}$
$=\text{e}\times\frac{\text{me}^4}{4\in_0^2\text{h}^3\text{n}^3}\times\big(\pi\text{r}^2_0\text{n}^2\big)=\frac{\text{me}^5\times\big(\pi\text{r}^2_0\text{n}^2\big)}{4\in_0^2\text{h}^3\text{n}^3}$
$=\frac{\big(9.1\times10^{-31}\big)\big(1.6\times10^{-19}\big)^5\times\pi\times(0.53)^2\times10^{-20}\times1}{4\times\big(8.85\times10^{-12}\big)^2\big(6.64\times10^{-34}\big)^3(1)^3}$
$=0.0009176\times10^{-20}=9176\times10^{-24}\text{Am}^2$

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Question 103 Marks

When a photon is emitted by a hydrogen atom, the photon carries a momentum with it. $(a)$ Calculate the momentum carries by the photon when a hydrogen atom emits light of wavelength $656.3\ nm . (b)$ With what speed does the atom recoil during this transition? Take the mass of the hydrogen atom $=1.67 \times 10^{-27} \mathrm{~kg}. (c)$ Find the kinetic energy of recoil of the atom.

Answer

$\lambda=656.3\text{nm}$

Momentum $\text{P}=\frac{\text{E}}{\text{C}}=\frac{\text{hc}}{\lambda}\times\frac{1}{\text{c}}=\frac{\text{h}}{\lambda}$
$\text{P}=\frac{6.63\times10^{-34}}{656.3\times10^{-9}}=0.01\times10^{-25}$

$\text{P}=1\times10^{-27}\text{kg-m/s}$

$1\times10^{-27}=1.67\times10^{-27}\times\text{v}$
$\text{v}=\frac{1}{1.67}=0.598=0.6\text{m/s}$

$KE$ of atom $=\frac{1}{2}\times1.67\times10^{-27}\times(0.6)^2$

$=\frac{0.3006\times10^{-27}}{1.6\times10^{-19}}\text{ev}$
$=1.9\times10^{-9}\text{ev}$

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Question 113 Marks

The numerical value of ionization energy in eV equals the ionization potential in volts. Does the equality hold if these quantities are measured in some other units?

Answer

The electron volt is the amount of energy given to an electron in order to move it through the electric potential difference of one volt.
$1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}$
The numerical value of ionisation energy in eV is equal to the ionisation potential in volts. The equality does not hold if these quantities are measured in some other units.

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Question 123 Marks

Radiation from hydrogen discharge tube falls on a cesium plate. Find the maximum possible kinetic energy of the photoelectrons. Work function of cesium is 1.9eV.

Answer

Wocs = 1.9eV
The radiations coming from the hydrogen discharge tube consist of photons of energy = 13.6eV.
Maximum KE of photoelectrons emitted
= Energy of Photons - Work function of metal.
= 13.6eV - 1.9eV = 11.7eV

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Question 133 Marks

Whenever a photon is emitted by hydrogen in Balmer series, it is followed by another photon in Lyman series. What wavelength does this latter photon correspond to?

Answer

The second wavelength is from Balmer to hyman i.e. from $n = 2$ to $n = 1, n_1 = 2$ to $n_2 = 1 \frac{1}{\lambda}=\text{R}\bigg(\frac{1}{\text{n}^2_1}-\frac{1}{\text{n}^2_2}\bigg)$
$\Rightarrow\frac{1}{\lambda}=1.097\times10^7\Big(\frac{1}{2^2}-\frac{1}{1^2}\Big)$
$\Rightarrow1.097\times10^7\Big(\frac{1}{4}-1\Big)$
$\Rightarrow\lambda=\frac{4}{1.097\times3}\times10^{-7}$
$\Rightarrow\lambda=1.215\times10^{-7}$
$\Rightarrow\lambda=121.5\times10^{-9}$
$\Rightarrow\lambda=122\text{nm}$

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Question 143 Marks

What is the energy of a hydrogen atom in the first excited state if the potential energy is taken to be zero in the ground state?

Answer

The potential energy of a hydrogen atom is zero in ground state. An electron is board to the nucleus with energy 13.6ev, Show we have to give energy of 13.6ev. To cancel that energy. Then additional 10.2ev. is required to attain first excited state. Total energy of an atom in the first excited state is = 13.6ev. + 10.2ev. = 23.8ev.

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Question 153 Marks

A hydrogen atom in ground state absorbs a photon of ultraviolet radiation of wavelength 50nm. Assuming that the entire photon energy is taken up by the electron with what kinetic energy will the electron be ejected?

Answer

$\lambda=50\text{nm}$Work function = Energy required to remove the electron from $\text{n}_1=1\text{ to }\text{n}_2=\infty$
$\text{E}=13.6\Big(\frac{1}{1}-\frac{1}{\infty}\Big)=13.6$
$\frac{\text{hc}}{\lambda}-13.6=\text{KE}$
$\frac{1242}{50}-13.6=\text{KE}$
$\text{KE}=24.84-13.6=11.24\text{eV}$

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Question 163 Marks

A positive ion having just one electron ejects it if a photon of wavelength $288\mathring{\text{A}}$ or less is absorbed by it. Identify the ion.

Answer

$\lambda=228\mathring{\text{A}}$$\text{E}=\frac{\text{hc}}{\lambda}=\frac{6.63\times10^{-34}\times3\times10^{8}}{228\times10^{-10}}=0.0872\times10^{-16}$
The transition takes place form n = 1 to n = 2
Now, ex. $\frac{13.6\times3}{4\times\text{z}^2}=0.0872\times10^{-16}$
$\Rightarrow\text{z}^2=\frac{0.0872\times10^{-16}}{13.6\times3\times1.6\times10^{-19}=5.3}$
$\Rightarrow\text{z}=\sqrt{5.3}=2.3$
The ion may be Helium.

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Question 173 Marks

Find the radius and energy of a $He^+$ ion in the states $(a) n = 1, (b) n = 4$ and $(c) n = 10.$

Answer

$\text{n}=1,\text{ r}=\frac{\in_0\text{h}^2\text{n}^2}{\pi\text{mZe}^2}=\frac{0.53\text{n}^2}{\text{Z}}\mathring{\text{A}}$

$=\frac{0.53\times1}{2}=0.265\mathring{\text{A}}$
$\in=\frac{-13.6\text{z}^2}{\text{n}^2}=\frac{-13.6\times4}{1}=-54.4\text{eV}$

$\text{n}=4,\text{ r}=\frac{0.53\times16}{2}=4.24\mathring{\text{A}}$

$\in=\frac{-13.6\times4}{164}=-3.4\text{eV}$

$\text{n}=10,\text{ r}=\frac{0.53\times100}{2}=26.5\mathring{\text{A}}$

$\in=\frac{-13.6\times4}{100}=-0.544\mathring{\text{A}}$

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Question 183 Marks

A hot gas emits radiation of wavelengths 46.0nm, 82.8nm and 103.5nm only. Assume that the atoms have only two excited states and the difference between consecutive energy levels decreases as energy is increased. Taking the energy of the highest energy state to be zero, find the energies of the ground state and the first excited state.

Answer

Energy in ground state is the energy acquired in the transition of $2^{nd}$ excited state to ground state. As $2^{nd}$ excited state is taken as zero level.$\text{E}=\frac{\text{hc}}{\lambda_1}=\frac{4.14\times10^{-15}\times3\times10^8}{46\times10^{-9}}=\frac{1242}{46}=27\text{ev}$
Again energy in the first excited state,$\text{E}=\frac{\text{hc}}{\lambda_1}=\frac{4.14\times10^{-15}\times3\times10^8}{103.5}=12\text{ev}$

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Question 193 Marks

A beam of light having wavelengths distributed uniformly between 450nm to 550nm passes through a sample of hydrogen gas. Which wavelength will have the least intensity in the transmitted beam?

Answer

The energies associated with 450nm radiation $=\frac{1242}{550}=2.258=2.26\text{ev}$ Energy associated with 550nm radiation $=\frac{1242}{550}=2.258=2.26\text{ev}$ The light comes under visible range, Thus, $\text{n}_1=2,\text{ n}_2=3,4,5,\ .....$$\text{E}_2-\text{E}_3=13.6\Big(\frac{1}{2^2}-\frac{1}{3^2}\Big)=1.9\text{ev}$
$\text{E}_2-\text{E}_4=13.6\Big(\frac{1}{4}-\frac{1}{16}\Big)=2.55\text{ev}$
$\text{E}_2-\text{E}_6=13.6\Big(\frac{1}{4}-\frac{1}{25}\Big)=2.856\text{ev}$
Only $E_2 - E_4$ comes in the range of energy provided. So the wavelength corresponding to that energy will be absorbed.$\lambda=\frac{1242}{2.55}=487.05\text{nm}=487\text{nm}$
487 nm wavelength will be absorbed.

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