Dual Nature of Matter JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Dual Nature of Matter, free to read — no sign-in needed. The full chapter has 24 questions; sign in to attempt the remaining 19 in the exam simulator.

Answers checked against the official answer keys.

1. Q1JEE Advanced Adv 2025 (Paper 1)
   Consider an electron in the $n=3$ orbit of a hydrogen-like atom with atomic number $Z$. At absolute temperature $T$, a neutron having thermal energy $k_{\mathrm{B}} T$ has the same de Broglie wavelength as that of this electron. If this temperature is given by $T=\frac{Z^2 h^2}{\alpha \pi^2 a_0^2 m_N k_B}$, (where $h$ is the Planck's constant, $k_B$ is the Boltzmann constant, $m_{\mathrm{N}}$ is the mass of the neutron and $a_0$ is the first Bohr radius of hydrogen atom) then the value of $\alpha$ is ________
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   Answer: 72

   $\frac{\mathrm{mv}^2}{\mathrm{r}}=\frac{\mathrm{KZe}^2}{\mathrm{r}^2}$ $\mathrm{mv}^2 \mathrm{r}=\frac{1}{4 \pi \epsilon_0} \mathrm{Ze}^2$ $\mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi}$ (1)/(2) gives $\mathrm{v}=\frac{\frac{\mathrm{Ze}^2}{4 \pi \epsilon_0}}{\frac{\mathrm{nh}}{2 \pi}}=\frac{\mathrm{Ze}^2}{2 \epsilon_0 \mathrm{nh}}$ $\begin{aligned} & \frac{h}{m v}=\frac{h}{\sqrt{2 m_N \cdot K_B T}} \\ & T=\frac{m^2 Z^2 e^4}{8 \epsilon_0^2 n^2 h^2 m_N K_B} \\ & n=3 \Rightarrow T=\frac{m^2 Z^2 e^4}{72 \epsilon_0^2 h^2 m_N K_B} \\ & \frac{(1)}{(2)^2} \Rightarrow \frac{1}{m r}=\frac{\frac{Z e^2}{4 \pi \epsilon_0}}{\frac{n^2 h^2}{4 \pi^2}}\end{aligned}$ $\begin{aligned} & \mathrm{r}=\frac{\mathrm{n}^2 \mathrm{~h}^2 \epsilon_0}{\pi \mathrm{Ze}^2 \cdot \mathrm{~m}} \Rightarrow \mathrm{a}_0=\frac{\mathrm{h}^2 \epsilon_0}{\pi \mathrm{e}^2 \mathrm{~m}} \\ & \mathrm{a}_0^2=\frac{\mathrm{h}^4 \epsilon_0^2}{\pi^2 \mathrm{e}^4 \mathrm{~m}^2} \\ & \mathrm{Ta}_0^2=\frac{\mathrm{m}^2 \mathrm{Z}^2 \mathrm{e}^4}{72 \epsilon_0 \mathrm{~h}^2 \mathrm{~m}_{\mathrm{N}} \mathrm{k}_{\mathrm{B}}} \cdot \frac{\mathrm{h}^4 \epsilon_0^2}{\pi^2 \mathrm{e}^4 \mathrm{~m}^2} \\ & \mathrm{~T}=\frac{\mathrm{h}^2 \mathrm{Z}^2}{72 \pi^2 \mathrm{a}_0^2 \mathrm{~m}_{\mathrm{N}} \mathrm{k}_{\mathrm{B}}} \Rightarrow \alpha=72\end{aligned}$
2. Q2JEE Advanced Adv 2023 (Paper 1)
   A Hydrogen-like atom has atomic number $Z$. Photons emitted in the electronic transitions from level $n=4$ to level $n=3$ in these atoms are used to perform photoelectric effect experiment on a target metal. The maximum kinetic energy of the photoelectrons generated is $1.95eV$. If the photoelectric threshold wavelength for the target metal is $310nm$, the value of $Z$ is _____. [Given $hc=1240eV-nm$ and $Rhc=13.6eV$, where $R$ is the Rydberg constant, $h$ is the Planck’s constant and $c$ is the speed of light in vacuum]
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   Answer: 3

   From Einstein's photoelectric effect, we can write $E=K{E}_{max}+\phi \Rightarrow E=K{E}_{max}+\dfrac{hc}{\lambda }$ $\Rightarrow ∆{E}_{4to3}=1.95eV+\dfrac{1240}{310}eV \Rightarrow ∆{E}_{4to3}=5.95eV$ Now, as we know for Hydrogen like atoms, we can write $∆E=13.6{Z}^{2}\left(\dfrac{1}{{\left({n}_{1}\right)}^{2}}-\dfrac{1}{{\left({n}_{2}\right)}^{2}}\right)$ Therefore,$\Rightarrow 13.6{Z}^{2}\left(\dfrac{1}{{3}^{2}}-\dfrac{1}{{4}^{2}}\right)=5.95$ $\Rightarrow 13.6{Z}^{2}\dfrac{7}{9\times 16}=5.95$ $\Rightarrow {Z}^{2}=\dfrac{5.95\times 9\times 16}{13.6\times 7}=9$ $\Rightarrow Z=3$
3. Q3JEE Advanced Adv 2022 (Paper 2)
   When light of a given wavelength is incident on a metallic surface, the minimum potential needed to stop the emitted photoelectrons is $6.0V$. This potential drops to $0.6V$ if another source with wavelength four times that of the first one and intensity half of the first one is used. What are the wavelength of the first source and the work function of the metal, respectively? [Take$\dfrac{hc}{e}=1.24\times {10}^{-6}J{mC}^{-1}$ .]
   1. A.$1.72\times {10}^{-7}m,1.20eV$
   2. B.$1.72\times {10}^{-7}m,5.60eV$
   3. C.$3.78\times {10}^{-7}m,5.60eV$
   4. D.$3.78\times {10}^{-7}m,1.20eV$
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   Answer: (A)

   According to the Einstein's equation of photoelectric effect for the first case, $\dfrac{hc}{\lambda }-\phi =6eV\cdots (i)$ And for the second case, $\dfrac{hc}{4\lambda }-\phi =0.6eV$ Therefore, $\dfrac{3hc}{4\lambda }=5.4eV$ $\Rightarrow \lambda =\dfrac{3hc}{4\times 5.4eV}=\dfrac{3\times 1.24\times {10}^{-6}}{4\times 5.4}$ $\Rightarrow \lambda =1.72\times {10}^{-7}m$ $\Rightarrow$ from equation $\left(i\right)$ $\dfrac{hc}{1.72\times {10}^{-7}}\times \dfrac{1}{1.6\times {10}^{-19}}-\phi =6eV$ $\dfrac{2\times {10}^{-25}}{2.75\times {10}^{-26}}-\phi =6$ $\Rightarrow \phi =\left(7.27-6\right)\cong 1.2eV$
4. Q4JEE Advanced Adv 2021 (Paper 2)
   In a photoemission experiment, the maximum kinetic energies of photoelectrons from metals $P,Q$ and $R$ are ${E}_{P},{E}_{Q}$ and ${E}_{R}$, respectively, and they are related by ${E}_{P}=2{E}_{Q}=2{E}_{R}$. In this experiment, the same source of monochromatic light is used for metals $P$ and $Q$ while a different source of monochromatic light is used for the metal $R$. The work functions for metals $P,Q$ and $R$ are $4.0eV,4.5eV$ and $5.5eV$, respectively. The energy of the incident photon used for metal $R$, in $eV$ is ___.
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   Answer: 6

   Let the maximum kinetic energies of photoelectrons be${E}_{P}=2{E}_{Q}=2{E}_{R}=2E$. The energy of incident photon on metal $P\text{and}Q$ be $U$ and for $R$ be $U'$. Photoelectric equation, ${E}_{ph}=\phi +K{E}_{\max }$ Now, using it for the three cases, $U=4+2E...\left(1\right) U=4.5+E...\left(2\right) U'=5.5+E...\left(3\right)$ From $\left(1\right)\text{and}\left(2\right)$, $E=0.5eV$ From $\left(3\right)$, energy of the incident photon for metal $R$, $U'=5.5+0.5=6eV$
5. Q5JEE Advanced Adv 2017 (Paper 2)
   A photoelectric material having work-function ${\phi }_{0}$ is illuminated with a light of wavelength $λ λ<hcϕ0$. The fastest photoelectron has a de Broglie wavelength ${\lambda }_{d}$. A change in wavelength of the incident light by $\Delta \lambda$ results in a change $\Delta {\lambda }_{d}$ in ${\lambda }_{d}$. The ratio $\dfrac{\Delta {\lambda }_{d}}{\Delta \lambda }$ is proportional to
   1. A.$\dfrac{{\lambda }_{d}^{3}}{{\lambda }^{2}}$
   2. B.$\dfrac{{\lambda }_{d}^{3}}{\lambda }$
   3. C.$\dfrac{{\lambda }_{d}^{2}}{{\lambda }^{2}}$
   4. D.$\dfrac{{\lambda }_{d}}{\lambda }$
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   Answer: (A)

   According to the equation of photoelectric effect, $K{E}_{max}=\dfrac{hc}{\lambda }-{\phi }_{0}$ Kinetic energy in terms of momentum is given as $KE=\dfrac{{p}^{2}}{2m}$ $\dfrac{{p}^{2}}{2m}=\dfrac{hc}{\lambda }-{\phi }_{0}$ $\dfrac{{\left(\dfrac{h}{{\lambda }_{d}}\right)}^{2}}{2m}=\dfrac{hc}{\lambda }-{\phi }_{0}$ Assuming small changes, differentiating both sides, $\dfrac{{h}^{2}}{2m}\left(-\dfrac{2d{\lambda }_{d}}{{\lambda }_{d}^{3}}\right)=-\dfrac{hc}{{\lambda }^{2}}d\lambda$ $\dfrac{d{\lambda }_{d}}{d\lambda }\propto \dfrac{{\lambda }_{d}^{3}}{{\lambda }^{2}}$