Modern Physics Previous Year Questions with Solutions

1.0Introduction

Modern Physics is essential for IIT-JEE, covering topics like the Photoelectric Effect, which shows light's particle nature, and Bohr’s Atomic Model, explaining electron behavior. It also explores Wave-Particle Duality and Heisenberg’s Uncertainty Principle, Nuclear Physics (radioactivity, nuclear reactions),Additionally, X-rays and Einstein’s Photoelectric Equation have real-world applications in imaging, communication, and space. To succeed, focus on core concepts, practice problems, and review past papers.

2.0Key Concepts to Remember

Dual Nature Of Light: Light does not have a definite nature; rather, its nature depends on its experimental phenomenon. This is known as the dual nature of light.

Work Function()-It is the minimum energy an electron requires to escape from a metal surface. It is measured in electron volts(eV).

Einstein’s Quantum Theory of Light:Light behaves as Quanta. These energy quanta are called Photons, and they have definite Energy and Momentum and travel with the speed of light in a vacuum.

Photoelectric Effect:It is a phenomenon of ejecting electrons by falling light of a suitable frequency or wavelength on a metal surface. Ejected electrons are called photoelectrons, and the current flowing due to these photoelectrons is called photoelectric current.

Saturation current :When all the photoelectrons emitted by the cathode reach the anode then current flowing in the circuit at that instant is known as saturation current; this is the maximum value of Photoelectric current.

Stopping Potential:The minimum magnitude of the negative potential of the anode concerning the cathode for which the current is zero is called the stopping potential or cut-off voltage; this voltage is independent of intensity.

3.0Important Formulas

Energy of Photon	$E=h\nu =\frac{hc}{\lambda }=\frac{12400}{\lambda }\text{ eV-A˚}$
Linear Momentum of Photon	$p=\frac{E}{c}=\frac{h\nu }{c}=\frac{h}{\lambda }$
Effective Mass of Photon	$m=\frac{E}{c^2}=\frac{hc}{c^2\lambda }=\frac{h}{c\lambda }$
Intensity of Light	$$\begin{gathered} I=\frac{E}{At}=\frac{P}{A} \\ I=\frac{N(h\nu )}{At}=\frac{n(h\nu )}{A}(\therefore n=\frac{N}{t},\text{no. of photons per sec.}) \end{gathered}$$
Radiation Force	$F=n\frac{2h}{\lambda }\text{where }n=\frac{P\lambda }{hc}$
Radiation Pressure	$P=\frac{F}{A}=\frac{2P}{cA}=\frac{2I}{c}$
Einstein Photoelectric Equation	$$\begin{gathered} h\nu =K.E_{\text{max}}+{\varphi }_0 \\ K.E_{\text{max}}=h\nu -{\varphi }_0 \end{gathered}$$
Quantum Efficiency	$$\begin{gathered} \text{Quantum Efficiency}(x)=\frac{\text{number of electrons emitted per second}}{\text{total number of photons incident per second}}=\frac{n_e}{n_{ph}} \\ \text{If quantum efficiency is }x\%\text{ then }\frac{n_e}{n_{ph}}=\frac{x}{100} \\ {n}_{ph}=(5\times 10^{24}{\text{J}}^{-1}\cdot {\text{m}}^{-1})P\lambda \end{gathered}$$
de-Broglie wavelength	$$\begin{gathered} \lambda =\frac{h}{p}=\frac{h}{mv}=\frac{h}{\sqrt{2mK}} \\ \lambda =\frac{h}{\sqrt{2mqV}} \end{gathered}$$
Bohr quantisation Condition	$mvr=\frac{nh}{2\pi }$
Size of Nucleus	$R=R_0{A}^{1/3}\text{where }{R}_0=\text{Empirical Constant}=1.2\text{fermi}$
Density of Nucleus	$\text{Density}=\frac{3m_n}{4\pi R_0^3}\approx 2.3\times 10^{17}{\text{kg/m}}^3$
Mass Defect	$\Delta m=\left[Z{m}_p+(A-Z)m_n\right]-M$
Nuclear Binding Energy	$E_b=\Delta m\cdot c^2$
Types of Process
Distance of Closest Approach	$r_0=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{(2ze{)}^2}{E_K}(E_K=K.Eof\alpha -particle)$
Radius of Orbits	$r_n=\frac{n^2{h}^2{\epsilon }_0}{\pi m{e}^2}$
Velocity of revolving electrons	$v=\frac{e^2}{2nh{\epsilon }_0}$
Frequency of electron in an orbit	$\upsilon =\frac{m{e}^4}{4{\epsilon }_0^2{h}^3{n}^3}$
Electron Energy	$E_n=-\frac{m{e}^4}{8n^2{h}^2{\epsilon }_0^2}$
Frequency of Emitted Radiations	$\bar{\nu }=\frac{1}{\lambda }=R\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$
Electron energy level in Hydrogen atom	$E_n=-\frac{13.6}{n^2}eV$
Spectral Series of Hydrogen Atom	$\bar{\nu }=\frac{1}{\lambda }=R\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$

4.0Past Year Questions with Solutions on Modern Physics: JEE (Mains)

Q-1.The radius of the third stationary orbit of electrons for Bohr's atom is R. The radius of fourth stationary orbit will be:

$(1)\frac{4}{3}R(2)\frac{16}{9}R(3)\frac{3}{4}R(4)\frac{9}{16}R$     

Solution: Ans(2)

$$\begin{gathered} r\propto \frac{n^2}{Z} \\ \frac{r_4}{r_3}=\frac{4^2}{3^2}=\frac{16}{9} \\ {r}_4=\frac{16}{9}R \end{gathered}$$

Q-2.The threshold frequency of a metal with work function 6.63 eV is :

$$\begin{gathered} [(1)]16\times 10^{15}Hz \\ [(2)]16\times 10^{15}Hz \\ [(3)]1.6\times 10^{12}Hz \\ [(4)]1.6\times 10^{15}Hz \end{gathered}$$

Solution: Ans(4)

$$\begin{gathered} {\varphi }_0=h{\nu }_0 \\ 6.63\times 1.6\times 10^{-19}=6.63\times 10^{-34}{v}_0 \\ {\nu }_0\frac{1.6\times 10^{-19}}{10^{-34}}=1.6\times 10^{15}Hz \end{gathered}$$

Q-3.If Rydberg’s constant is R, the longest wavelength of radiation in Paschen series will be $\frac{\alpha }{7R}$ , where   = ________.

Solution: Ans $(\alpha =144)$

Longest wavelength corresponds to transition between n = 3 and n = 4  

$$\begin{gathered} \frac{1}{\lambda }=R{Z}^2\left(\frac{1}{3^2}-\frac{1}{4^2}\right)=R{Z}^2\left(\frac{1}{9}-\frac{1}{16}\right)=\frac{7R{Z}^2}{9\times 16}\Rightarrow \lambda =\frac{144}{7R}\text{for }Z=1 \\ \alpha =144 \end{gathered}$$

Q-4.The de-Broglie wavelength of an electron is the same as that of a photon. If velocity of electron is 25% of the velocity of light, then the ratio of K.E. of electron and K.E. of photon will be:

$[(1)]\frac{1}{1}[(2)]\frac{1}{8}[(3)]\frac{8}{1}[(4)]\frac{1}{4}$

Solution: (2)

$$\begin{gathered} \text{For Photon} \\ {E}_p=\frac{hc}{{\lambda }_p}\Rightarrow {\lambda }_p=\frac{hc}{E_p} \\ \text{For Electron} \\ {\lambda }_p=\frac{h}{m_e{v}_e}=\frac{h{v}_e}{2K_e} \\ \text{Given}{v}_e=0.25c \\ {\lambda }_e=\frac{h\times 0.25c}{2K_e}=\frac{hc}{8K_e} \\ Also,{\lambda }_p={\lambda }_e\frac{hc}{E_p}=\frac{hc}{8K_e}\Rightarrow \frac{K_e}{E_p}=\frac{1}{8} \end{gathered}$$

Q-5.The explosive in a Hydrogen bomb is a mixture of ₁H², ₁H³ and ₃Li⁶ in some condensed form. The chain reaction is given by

₃Li⁶ + ₀n¹₂He⁴ + ₁H³

₁H² + ₁H³₂He⁴ + ₀n¹

During the explosion the energy released is approximately

[Given : M(Li) = 6.01690 amu. M (₁H²) = 2.01471 amu. M (₂He⁴) = 4.00388amu, and 1 amu = 931.5 MeV]

(1) 28.12 MeV           (2) 12.64 MeV            (3) 16.48 MeV                (4) 22.22 MeV

Solution: Ans(4)

$Q=\Delta m{c}^2$

Q = [M(Li)+ M (₁H²) –2 × M(₂He⁴)] × 931.5 MeV

Q = [6.01690+2.01471–2 × 4.00388] × 931.5 MeV

Q = 22.216 MeV

Q = 22.22 MeV

Q-6.When a hydrogen atom going from n = 2 to n = 1 emits a photon, its recoil speed is x5 m/s  \frac{x}{5} m/s Where  x = ______ . (Use : mass of hydrogen atom
= 1.6 × 10^–27 kg)

Solution:Ans(x=17)

Recoil Speed

$$\begin{gathered} V=\frac{\Delta E}{mc} \\ V=\frac{10.2eV}{1.6\times 10^{-27}\times 3\times 10^8} \\ V=\frac{10.2\times 1.6\times 10^{-19}}{1.6\times 10^{-27}\times 3\times 10^8} \\ V=3.4m/s=\frac{17}{5}m/s\Rightarrow x=17 \end{gathered}$$

Q-7.Two sources of light emit with a power of 200 W. The ratio of number of photons of visible light emitted by each source having wavelengths 300 nm and 500 nm respectively, will be :

(1) 1 : 5             (2) 1 : 3                (3) 5 : 3                (4) 3 : 5

Solution: Ans(4)

$$\begin{gathered} n_1\times \frac{hc}{{\lambda }_1}=200 \\ {n}_2\times \frac{hc}{{\lambda }_2}=200 \\ \frac{n_1}{n_2}=\frac{{\lambda }_1}{{\lambda }_2}=\frac{300}{500} \\ \Rightarrow \frac{n_1}{n_2}=\frac{3}{5} \end{gathered}$$

Q-8.Hydrogen atom is bombarded with electrons accelerated through a potential difference of V, which causes excitation of hydrogen atoms. If the experiment is being formed at T = 0 K. The minimum potential difference needed to observe any Balmer series lines in the emission spectra will be $\frac{\alpha }{10}$, where = ________. $\alpha$ =

Solution: Ans(121)

For minimum potential difference electron has to make transition from n=3 to n=2 state but first electron has to reach n=3 state from ground state. So energy bombarding electron should be equal to energy difference of n=3 and n=1 state.

$$\begin{gathered} \Delta E=13.6\left[1-\frac{1}{3^2}\right]=eV \\ \Rightarrow \frac{13.6\times 8}{9}=V \\ V=12.09 \\ V\approx 12.1V \\ \alpha =121 \end{gathered}$$

Q-9.The work function of a substance is 3.0 eV. The longest wavelength of light that can cause the emission of photoelectrons from this substance is approximately:

(1) 215 nm             (2) 414 nm                (3) 400 nm            (4) 200 nm

Solution: Ans(2)

$$\begin{gathered} \text{Work Function}=3eV \\ \frac{hc}{{\lambda }_{th}}=3eV\text{or}{\lambda }_{th}=\frac{12420}{3}\text{A˚} \\ =414nm\text{(Approx)} \end{gathered}$$

Q-10.A electron of a hydrogen atom in an excited state is having energy E_n = – 0.85 eV. The maximum number of allowed transitions to lower energy level is ….

Solution: Ans(6)

$$\begin{gathered} E_n=-\frac{13.6}{n^2}=0.85 \\ \Rightarrow n=4 \\ \text{No. of Transitions}=\frac{n(n-1)}{2}=\frac{4(4-1)}{2}=6 \end{gathered}$$

Q-11.For the photoelectric effect, the maximum kinetic energy (E_k) of the photoelectrons is plotted against the frequency (v) of the incident photons as shown in the figure. The slope of the graph gives

(1) Ratio of Planck’s constant to electric charge

(2) Work function of the metal

(3) Charge of electron

(4) Planck’s constant

Solution: Ans(4)

$$\begin{gathered} K.E=h\nu -\varphi \\ \tan \theta =h \end{gathered}$$

Q-12.An electron revolving in n^th Bohr orbit has magnetic moment n . If n nx , the value of x is:

(1) 2            (2) 1              (3) 3                (4) 0 

Solution: Ans(2)

$$\begin{gathered} \text{magnetic moment}=i\pi r^2 \\ \mu =\frac{evr}{2} \\ \mu \propto \left(\frac{1}{n}\right)n^2\Rightarrow \mu \propto n \\ x=1 \end{gathered}$$

Q-13.If the total energy transferred to a surface in time t is 6.48 × 10⁵ J, then the magnitude of the total momentum delivered to this surface for complete absorption will be :

(1) 2.46 × 10^–3 kg m/s

(2) 2.16 × 10^–3 kg m/s

(3) 1.58 × 10^–3 kg m/s

(4) 4.32 × 10^–3 kg m/s

Solution: Ans(2)

$p=\frac{E}{c}=\frac{6.48\times 10^5}{3\times 10^8}=2.16\times 10^{-3}$

Q-14.If the wavelength of the first member of the Lyman series of hydrogen is l. The wavelength of the second member will be

$$\begin{gathered} (1)\frac{27}{32}\lambda \\ (2)\frac{32}{27}\lambda \\ (3)\frac{27}{5}\lambda \\ (4)\frac{5}{27}\lambda \end{gathered}$$

Solution:Ans(1)

$$\begin{gathered} \frac{1}{\lambda }=\frac{13.6Z^2}{hc}\left[1-\frac{1}{2^2}\right]\text{……(1)} \\ \frac{1}{{\lambda }^{\prime }}=\frac{13.6Z^2}{hc}\left[1-\frac{1}{3^2}\right]\text{……(2)} \\ \text{On dividing (1) and (2)} \\ {\lambda }^{\prime }=\frac{27}{32}\lambda \end{gathered}$$

Q-15.When a metal surface is illuminated by light of wavelength l, the stopping potential is 8V. When the same surface is illuminated by light of wavelength 3l, stopping potential is 2V. The threshold wavelength for this surface is :

$$\begin{gathered} (1)5\lambda \\ (2)3\lambda \\ (3)9\lambda \\ (4)4.5\lambda \end{gathered}$$

Solution: Ans(3)

$$\begin{gathered} E=\varphi +K_{\text{max}} \\ \varphi =\frac{hc}{{\lambda }_0} \\ {K}_{\text{max}}=e{V}_0 \\ 8e=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0}\text{……(1)} \\ 2e=\frac{hc}{3\lambda }-\frac{hc}{{\lambda }_0}\text{……(2)} \\ \text{On solving (1) and (2)}{\lambda }_0=9\lambda \end{gathered}$$

Q-16.A particular hydrogen - like ion emits the radiation of frequency 3 × 10¹⁵ Hz when it makes transition from n = 2 to n = l.  The frequency of radiation emitted in transition from n = 3 to n = l is $\frac{x}{9}\times 10^{15}hz$ when x = ______

Solution: Ans(x=32)

$$\begin{gathered} E=13.6\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \\ \text{Energy released in transition} \\ {E}_1=13.6\left(\frac{1}{1}-\frac{1}{4}\right)eV \\ h{f}_1=13.6\times \frac{3}{4}eV\text{……(1)} \\ {E}_2=13.6\times \frac{8}{9}eV \\ h{f}_2=13.6\times \frac{8}{9}eV\text{……(2)} \\ \text{(2) divided by (1)}\frac{f_2}{f_1}=\frac{13.6\times \frac{8}{9}}{13.6\times \frac{3}{4}}=\frac{32}{27} \\ {f}_2=\frac{32}{27}\times f_1=\frac{32}{27}\times 3\times 10^{15}=\frac{32}{9}\times 10^{15}\text{Hz} \\ {f}_2=\frac{x}{9}\times 10^{15} \\ x=32 \end{gathered}$$

Q-17.The radius of a nucleus of mass number 64 is 4.8 fermi. Then the mass number of another nucleus having radius of 4 fermi is $\frac{1000}{x}$, where x is _____.

Solution: Ans(x=27)

$$\begin{gathered} R=R_0{A}^{1/3} \\ {R}^3\propto A \\ {\left(\frac{4.8}{4}\right)}^3=\frac{64}{A} \\ (1.2{)}^3=\frac{64}{A} \\ A=\frac{64}{1.44\times 1.2}=\frac{1000}{x} \\ x=\frac{144\times 12}{64}=27 \end{gathered}$$

Q-18.The mass number of nucleus having radius equal to half of the radius of nucleus with mass number 192 is:

(1) 24             (2) 32             (3) 40                (4) 20 

Solution: (1)

$$\begin{gathered} R_1=\frac{R_2}{2} \\ {R}_0{\left(A_1\right)}^{1/3}=\frac{R_0}{2}{\left(A_2\right)}^{1/3} \\ {A}_1=\frac{1}{8}{A}_2=\frac{192}{8}=24 \end{gathered}$$