Download App

Physics 2 Marks Questions

Bohr’s Model and Physics of Atom · Gujarat Board (GSEB) STD 12 Science (GSEB - English Medium). 17 questions.

Questions

2 Marks Questions

🎯

Test yourself on this topic

17 questions · timed · auto-graded

Question 12 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
n1 = 2, n2 = 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.
View full question & answer→
Question 22 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 × 10-9C
Let V0 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
$\frac{4.14\times10^{-15}\times3\times10^8}{440\times10^{-9}}-2\text{eV}=\text{eV}_0$
$\text{eV}_0=\Big(\frac{1242}{440}-2\Big)\text{eV}$
$\text{eV}_0=0.823\text{eV}$
$\text{V}_0=0.823\text{ volts}$ 
View full question & answer→
Question 32 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 × 10-5eVK-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 H2+ which is impossible.
View full question & answer→
Question 42 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 $\frac{\text{V}_0}{2\pi\text{r}_0}$
[r0 = radius of ground state, V0 = velocity in the ground state]
$\therefore$ Frequency of radiation emitted is $\frac{\text{V}_0}{2\pi\text{r}_0}=\text{f}$
$\therefore\text{C}=\text{f}\lambda$
$\lambda=\frac{\text{C}}{\text{f}}=\frac{2\pi\text{C}\text{r}_0}{\text{V}_0}$
$\therefore\lambda=\frac{2\pi\text{C}\text{r}_0}{\text{V}_0}$
$\lambda=45.686\text{nm}=45.7\text{nm}$
View full question & answer→
Question 52 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]
$=\frac{\big(1.6\times10^{-19}\big)\times\big(1.6\times10^{-19}\big)\times9\times10^9}{\big(0.53\times10^{-10}\big)^2}$
$=82.02\times10^{-9}=8.202\times10^{-8}=8.2\times10^{-8}\text{N}$
View full question & answer→
Question 62 Marks
Find the binding energy of a hydrogen atom in the state n = 2.
Answer
$\text{n}_1=2,\ \text{n}_2=\infty$
$\text{E}=\frac{-13.6}{\text{n}^2_1}-\frac{-13.6}{\text{n}^2_2}$
$\text{E}=13.6\bigg(\frac{1}{\text{n}^2_1}-\frac{1}{\text{n}^2_2}\bigg)$
$\text{E}=13.6\Big(\frac{1}{\infty}-\frac{1}{4}\Big)$
$\text{E}=\frac{-13.6}{4}=-3.4\text{eV}$
View full question & answer→
Question 72 Marks
Find the maximum angular speed of the electron of a hydrogen atom in a stationary orbit.
Answer
We know, $\text{m}\mu\text{r}=\frac{\text{nh}}{2\pi}$
$\Rightarrow\text{mr}^2\text{w}=\frac{\text{nh}}{2\pi}$
$\Rightarrow\text{w}=\frac{\text{hn}}{2\pi\times\text{m}\times\text{r}^2}$
$\Rightarrow\text{w}=\frac{1\times6.63\times10^{-34}}{2\times3.14\times9.1\times10^{-31}\times(0.53)^2\times10^{-20}}$
$\Rightarrow\text{w}=0.413\times10^{17}\text{rad/s}$
$\Rightarrow\text{w}=4.13\times10^{17}\text{rad/s}$ 
View full question & answer→
Question 82 Marks
Evaluate Rydberg constant by putting the values of the fundamental constants in its expression.
Answer
$\text{Rydbergs constant}=\frac{\text{me}^4}{8\text{h}^3\text{C}\in_0^2}$
$\text{m}_\text{e}=9.1\times10^{-31}\text{kg},\text{ e}=1.6\times10^{-19}\text{c},\text{ h}=6.63\times10^{-34}\text{J-s}$
$\text{C}=3\times10^{8}\text{m/s},\in_0=8.85\times10^{-12}$
$\text{R}=\frac{9.1\times10^{-31}\times(1.6\times10^{-19})^4}{8\times(6.63\times10^{-34})^3\times3\times10^8\times(8.85\times10^{-12})^2}$
$\text{R}=1.097\times10^{7}\text{m}^{-1}$
View full question & answer→
Question 92 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.
View full question & answer→
Question 102 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 n1 = 3 to n2 = 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
View full question & answer→
Question 112 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]
View full question & answer→
Question 122 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 n2 = 3).
View full question & answer→
Question 132 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.
View full question & answer→
Question 142 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
  1. $\text{n}_1=1,\text{ n}_2=3,\text{ E}=13.6\Big(\frac{1}{1}-\frac{1}{9}\Big)=\frac{13.6\times8}{9}=\frac{\text{hc}}{\lambda}$

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

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

  1. 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 $\lambda=487\text{nm}$

View full question & answer→
Question 152 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
  1. The gas emits 6 wavelengths, let it be in nth excited state.

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

$\Rightarrow\text{n}=4$

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

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

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

$\Rightarrow\text{n}=4$

View full question & answer→
Question 162 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
  1. 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

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

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

View full question & answer→
Question 172 Marks
  1. Find the first excitation potential of He+ ion.
  2. Find the ionization potential of Li++ ion.
Answer
  1. First excitation potential of,

He+ = 10.2 × z2 = 10.2 × 4 = 40.8V

  1. Ionization potential of L1++

= 13.6V × z2 = 13.6 × 9 = 122.4V

View full question & answer→
Generate Paper FreeAll Bohr’s Model and Physics of Atom topics