2 Marks Questions - Physics STD 12 Science Questions - Vidyadip

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2 Marks Questions

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

Question 12 Marks

A spectroscopic instrument can resolve two nearby wavelengths $\lambda$ and $\lambda+\Delta\lambda$ if $\frac{\lambda}{\Delta\lambda}$ is smaller than 8000. This is used to study the spectral lines of the Balmer series of hydrogen. Approximately how many lines will be resolved by the instrument?

Answer

The range of Balmer series is 656.3nm to 365nm. It can resolve $\lambda$ and $\lambda+\Delta\lambda$ if $\frac{\lambda}{\Delta\lambda}=8000$
$\therefore$ No. of wavelength in the range $=\frac{656.3-365}{8000}=36$
Total no. of lines 36 + 2 = 38 [extra two is for first and last wavelength]

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Question 22 Marks

Suppose, in certain conditions only those transitions are allowed to hydrogen atoms in which the principal quantum number n changes by 2. (a) Find the smallest wavelength emitted by hydrogen. (b) List the wavelength emitted by hydrogen in the visible range (380nm to 780nm).

Answer

$\text{n}_1=1,\text{ n}_2=3,\text{ E}=13.6\Big(\frac{1}{1}-\frac{1}{9}\Big)=\frac{13.6\times8}{9}=\frac{\text{hc}}{\lambda}$

or $\frac{13.6\times8}{9}=\frac{4.14\times10^{-15}\times3\times10^8}{\lambda}$

$\lambda=\frac{4.14\ \times\ 3\ \times\ 10^{-7}}{13.6\ \times\ 8}=1.027\times10^{-7}=103\text{nm}$

As ‘n’ changes by 2, we may consider n = 2 to n = 4

then, $\text{E}=13.6\times\Big(\frac{1}{4}-\frac{1}{16}\Big)=2.55\text{ev}$ and $2.55=\frac{1242}{\lambda}$ or $\lambda=487\text{nm}$

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Question 32 Marks

The light emitted in the transition n = 3 to n = 2 in hydrogen is called $\text{H}_\alpha$ light. Find the maximum work function a metal can have so that $\text{H}_\alpha$ light can emit photoelectrons from it.

Answer

$n_1 = 2, n_2 = 3$
Energy possessed by $\text{H}_\alpha$ light
$=13.6\bigg(\frac{1}{\text{n}^2_1}-\frac{1}{\text{n}^2_2}\Big)$
$=13.6\times\Big(\frac{1}{4}-\frac{1}{9}\Big)$
$=1.89\text{ev}$
For $\text{H}_\alpha$ light to be able to emit photoelectrons from a metal the work function must be greater than or equal to 1.89ev.

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Question 42 Marks

Find the maximum Coulomb force that can act on the electron due to the nucleus in a hydrogen atom.

Answer

$\text{F}=\frac{\text{q}_1\text{q}_2}{4\pi\in_0\text{r}^2}$
[Smallest dist. Between the electron and nucleus in the radius of first Bohrs orbit]
$=\frac{\big(1.6\times10^{-19}\big)\times\big(1.6\times10^{-19}\big)\times9\times10^9}{\big(0.53\times10^{-10}\big)^2}$
$=82.02\times10^{-9}=8.202\times10^{-8}=8.2\times10^{-8}\text{N}$

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Question 52 Marks

A filter transmits only the radiation of wavelength greater than 440nm. Radiation from a hydrogen-discharge tube goes through such a filter and is incident on a metal of work function 2.0eV. Find the stopping potential which can stop the photoelectrons.

Answer

Wavelength of radiation coming from filter, $\lambda=440\text{nm}$
Work function of metal, $\phi=2\text{eV}$
Charge of an electron $e = 1.6 \times 10^{-9}C$
Let $V_0$ be the_stopping potential.
From Einstein's photoelectric equation,
$\frac{\text{hc}}{\lambda}-\phi=\text{eV}_0$
Here,
h =Planck constant
c = Speed of light
$\lambda=$ Wavelength of radiation
$\frac{4.14\times10^{-15}\times3\times10^8}{440\times10^{-9}}-2\text{eV}=\text{eV}_0$
$\text{eV}_0=\Big(\frac{1242}{440}-2\Big)\text{eV}$
$\text{eV}_0=0.823\text{eV}$
$\text{V}_0=0.823\text{ volts}$

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Question 62 Marks

The average kinetic energy of molecules in a gas at temperature T is 1.5kT. Find the temperature at which the average kinetic energy of the molecules of hydrogen equals the binding energy of its atoms. Will hydrogen remain in molecular from at this temperature? Take $k = 8.62 \times 10^{-5}eVK^{-1}.$

Answer

$\text{KE}=\frac{3}{2}\text{KT}=1.5\text{KT},\text{ K}=8.62\times10^{-5}\text{eV/k},$
$\text{Binding Energy}=-13.6\Big(\frac{1}{\infty}-\frac{1}{1}\Big)=13.6\text{eV}$
According to the question, 1.5KT = 13.6
$\Rightarrow1.5\times8.62\times10^{-5}\times\text{T}=13.6$
$\Rightarrow\text{T}=\frac{13.6}{1.5\ \times\ 8.62\ \times\ 10^{-5}}=1.05\times10^5\text{K}$
No, because the molecule exists an ${H_2}^+$ which is impossible.

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Question 72 Marks

According to Maxwell's theory of electrodynamics, an electron going in a circle should emit radiation of frequency equal to its frequency of revolution. What should be the wavelength of the radiation emitted by a hydrogen atom in ground state if this rule is followed?

Answer

Frequency of the revolution in the ground state is $\frac{\text{V}_0}{2\pi\text{r}_0}$
$[r_0 =$ radius of ground state, $V_0 =$ velocity in the ground state$]$
$\therefore$ Frequency of radiation emitted is $\frac{\text{V}_0}{2\pi\text{r}_0}=\text{f}$
$\therefore\text{C}=\text{f}\lambda$
$\lambda=\frac{\text{C}}{\text{f}}=\frac{2\pi\text{C}\text{r}_0}{\text{V}_0}$
$\therefore\lambda=\frac{2\pi\text{C}\text{r}_0}{\text{V}_0}$
$\lambda=45.686\text{nm}=45.7\text{nm}$

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Question 82 Marks

A gas of hydrogen-like ions is prepared in a particular excited state A. It emits photons having wavelength equal to the wavelength of the first line of the Lyman series together with photons of five other wavelengths. Identify the gas and find the principal quantum number of the state A.

Answer

The gas emits 6 wavelengths, let it be in $n^{th}$ excited state.

$\Rightarrow\frac{\text{n}(\text{n}-1)}{2}=6$

$\Rightarrow\text{n}=4$

$\therefore$ The gas is in $4^{th}$ excited state.

Total no. of wavelengths in the transition is 6. We have,

$\frac{\text{n}(\text{n}-1)}{2}=6$

$\Rightarrow\text{n}=4$

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Question 92 Marks

Balmer series was observed and analysed before the other series. Can you suggest a reason for such an order?

Answer

The Balmer series lies in the visible range. Therefore, it was observed and analysed before the other series. The wavelength range of Balmer series is from 364nm $(\text{for n}_2=\infty)$ to 655nm (for $n_2= 3$).

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Question 102 Marks

Find the first excitation potential of $He^+$ ion.
Find the ionization potential of $Li^{++}$ion.

Answer

First excitation potential of,

$He^+ = 10.2 \times z^2 = 10.2 \times 4 = 40.8V$

Ionization potential of $ L_1^{++}$

$= 13.6V \times z^2 = 13.6 \times 9 = 122.4V$

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Question 112 Marks

Light from Balmer series of hydrogen is able to eject photoelectrons from a metal. What can be the maximum work function of the metal?

Answer

The maximum energy liberated by the Balmer Series is $\text{n}_1=2,\text{ n}_2=\infty$
$\text{E}=13.6\bigg(\frac{1}{\text{n}^2_1}-\frac{1}{\text{n}^2_2}\bigg)=13.6\times\frac{1}{4}=3.4\text{ev}$
3.4ev is the maximum work function of the metal.

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Question 122 Marks

Electrons are emitted from an electron gun at almost zero velocity and are accelerated by an electric field E through a distance of 1.0m. The electrons are now scattered by an atomic hydrogen sample in ground state. What should be the minimum value of E so that red light of wavelength 656.3nm may be emitted by the hydrogen?

Answer

The given wavelength in Balmer series.
The first line, which requires minimum energy is from $n_1 = 3$ to $n_2= 2$
$\therefore$ The energy should be equal to the energy required for transition from ground state to $n = 3$
i.e., $\text{E}=13.6\Big[1-\Big(\frac{1}{9}\Big)\Big]=12.09\text{eV}$
$\therefore$ Minimum value of electric field $= 12.09v/m = 12.1v/m$

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Question 132 Marks

A hydrogen atom in a state having a binding energy of 0.85eV makes transition to a state with excitation energy 10.2eV (a) Identify the quantum numbers n of the upper and the lower energy states involved in the transition. (b) Find the wavelength of the emitted radiation.

Answer

From the energy data we see that the H atom transists from binding energy of 0.85 eV to exitation energy of 10.2eV = Binding Energy of -3.4eV.

So, n = 4 to n = 2

We know $=\frac{1}{\lambda}=1.097\times10^{7}\Big(\frac{1}{4}-\frac{1}{16}\Big)$

$\lambda=\frac{16}{1.097\times3\times10^7}=4.8617\times10^{-7}=487\text{nm}$

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Question 142 Marks

A neutron having kinetic energy 12.5eV collides with a hydrogen atom at rest. Nelgect the difference in mass between the neutron and the hydrogen atom and assume that the neutron does not leave its line of motion. Find the possible kinetic energies of the neutron after the event.

Answer

In one dimensional elastic collision of two bodies of equal masses. The initial velocities of bodies are interchanged after collision.
$\therefore$ Velocity of the neutron after collision is zero.
Hence, it has zero energy.

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Question 152 Marks

Find the binding energy of a hydrogen atom in the state n = 2.

Answer

$\text{n}_1=2,\ \text{n}_2=\infty$
$\text{E}=\frac{-13.6}{\text{n}^2_1}-\frac{-13.6}{\text{n}^2_2}$
$\text{E}=13.6\bigg(\frac{1}{\text{n}^2_1}-\frac{1}{\text{n}^2_2}\bigg)$
$\text{E}=13.6\Big(\frac{1}{\infty}-\frac{1}{4}\Big)$
$\text{E}=\frac{-13.6}{4}=-3.4\text{eV}$

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Question 162 Marks

Find the maximum angular speed of the electron of a hydrogen atom in a stationary orbit.

Answer

We know, $\text{m}\mu\text{r}=\frac{\text{nh}}{2\pi}$
$\Rightarrow\text{mr}^2\text{w}=\frac{\text{nh}}{2\pi}$
$\Rightarrow\text{w}=\frac{\text{hn}}{2\pi\times\text{m}\times\text{r}^2}$
$\Rightarrow\text{w}=\frac{1\times6.63\times10^{-34}}{2\times3.14\times9.1\times10^{-31}\times(0.53)^2\times10^{-20}}$
$\Rightarrow\text{w}=0.413\times10^{17}\text{rad/s}$
$\Rightarrow\text{w}=4.13\times10^{17}\text{rad/s}$

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Question 172 Marks

Evaluate Rydberg constant by putting the values of the fundamental constants in its expression.

Answer

$\text{Rydbergs constant}=\frac{\text{me}^4}{8\text{h}^3\text{C}\in_0^2}$
$\text{m}_\text{e}=9.1\times10^{-31}\text{kg},\text{ e}=1.6\times10^{-19}\text{c},\text{ h}=6.63\times10^{-34}\text{J-s}$
$\text{C}=3\times10^{8}\text{m/s},\in_0=8.85\times10^{-12}$
$\text{R}=\frac{9.1\times10^{-31}\times(1.6\times10^{-19})^4}{8\times(6.63\times10^{-34})^3\times3\times10^8\times(8.85\times10^{-12})^2}$
$\text{R}=1.097\times10^{7}\text{m}^{-1}$

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