Dual Nature of Radiation & Matter — Previous Year Questions (JEE Main 2026)
Solved JEE Main 2026 previous year questions on Dual Nature of Radiation and Matter — photoelectric effect, work function, stopping potential and de Broglie wavelength — with step-by-step solutions.
Solved JEE Main 2026 questions from the Dual Nature of Radiation and Matter chapter, covering the photoelectric effect, work function, stopping potential, and the de Broglie wavelength, each with a concise step-by-step solution.
Solutions are AI-generated and pending review.
Solution
Einstein’s photoelectric equation: $eV_s = \dfrac{hc}{\lambda} - \phi$.
Case 1 (wavelength $\lambda$, stopping potential $3V_o$):
$$3eV_o = \frac{hc}{\lambda} - \phi$$Case 2 (wavelength $2\lambda$, stopping potential $V_o$):
$$eV_o = \frac{hc}{2\lambda} - \phi$$Subtract:
$$2eV_o = \frac{hc}{\lambda} - \frac{hc}{2\lambda} = \frac{hc}{2\lambda} \implies eV_o = \frac{hc}{4\lambda}$$Work function (from Case 2):
$$\phi = \frac{hc}{2\lambda} - eV_o = \frac{hc}{2\lambda} - \frac{hc}{4\lambda} = \frac{hc}{4\lambda}$$Threshold wavelength: $\phi = \dfrac{hc}{\lambda_0}$
$$\lambda_0 = \frac{hc}{\phi} = \frac{hc}{hc/4\lambda} = 4\lambda \implies \alpha = 4$$Answer: B ($\alpha = 4$)
Solution
A. Planck’s constant — $E = h\nu \Rightarrow h = \dfrac{E}{\nu}$, units $\mathrm{J\,s}$:
$$[h] = \frac{\mathrm{M\,L^2\,T^{-2}}}{\mathrm{T^{-1}}} = \mathrm{M\,L^2\,T^{-1}} \;\to\; \text{III}$$B. Stopping potential — a potential (volt) $= \dfrac{\text{energy}}{\text{charge}}$:
$$[V] = \frac{\mathrm{M\,L^2\,T^{-2}}}{\mathrm{A\,T}} = \mathrm{M\,L^2\,T^{-3}\,A^{-1}} \;\to\; \text{IV}$$C. Work function — an energy:
$$[\phi] = \mathrm{M\,L^2\,T^{-2}} \;\to\; \text{I}$$D. Threshold frequency — a frequency:
$$[\nu_0] = \mathrm{T^{-1}} \;\to\; \text{II}$$Hence A-III, B-IV, C-I, D-II.
Answer: A
Solution
A charge $q$ accelerated through $V$ gains kinetic energy $qV = \dfrac{p^2}{2m}$, so momentum $p = \sqrt{2mqV}$.
$$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mqV}}$$The electron and proton carry the same charge magnitude $e$ and are accelerated through the same $V$:
$$\frac{\lambda_e}{\lambda_p} = \frac{h/\sqrt{2m_e eV}}{h/\sqrt{2m_p eV}} = \sqrt{\frac{m_p}{m_e}}$$Answer: A $\left(\sqrt{\dfrac{m_p}{m_e}}\right)$
Solution
For an electron accelerated through $V$:
$$\lambda = \frac{h}{\sqrt{2meV}} \implies \lambda \propto \frac{1}{\sqrt{V}}$$A 50% increase means $\lambda_2 = 1.5\,\lambda_1$:
$$\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{V_2}} = 1.5$$Square both sides:
$$\frac{V_1}{V_2} = (1.5)^2 = 2.25 = \frac{9}{4}$$Comparing with $\dfrac{V_1}{V_2} = \dfrac{9}{\alpha}$ gives $\alpha = 4$.
Answer: 4
Solution
Force on the electron (charge $-e$):
$$\vec{F} = (-e)\vec{E} = (-e)(-2E_0\hat{i}) = 2eE_0\,\hat{i}$$The force is along $+\hat{i}$, the same direction as the initial velocity, so the electron speeds up.
Acceleration:
$$a = \frac{2eE_0}{m}$$Velocity at time $t$:
$$v = v_0 + at = v_0 + \frac{2eE_0}{m}t = v_0\left(1 + \frac{2E_0 e}{m}\frac{t}{v_0}\right)$$de Broglie wavelength: $\lambda = \dfrac{h}{mv}$. Using $\lambda_0 = \dfrac{h}{4mv_0} \Rightarrow \dfrac{h}{m} = 4\lambda_0 v_0$:
$$\lambda = \frac{h/m}{v} = \frac{4\lambda_0 v_0}{v_0\left(1 + \dfrac{2E_0 e}{m}\dfrac{t}{v_0}\right)} = \frac{4\lambda_0}{\left[1 + \dfrac{2E_0 e}{m}\dfrac{t}{v_0}\right]}$$Answer: C
Solution
Einstein’s equation: $\phi = \dfrac{hc}{\lambda} - eV_s$.
Photon energy:
$$\frac{hc}{\lambda} = \frac{(6.62\times10^{-34})(3\times10^{8})}{331\times10^{-9}} = \frac{1.986\times10^{-25}}{3.31\times10^{-7}} = 6.0\times10^{-19}\ \text{J}$$Energy carried by stopping potential:
$$eV_s = (1.6\times10^{-19})(0.2) = 0.32\times10^{-19}\ \text{J}$$Work function:
$$\phi = 6.0\times10^{-19} - 0.32\times10^{-19} = 5.68\times10^{-19}\ \text{J}$$So $\alpha = 5.68$.
Answer: C ($5.68$)
Solution
de Broglie wavelength: $\lambda = \dfrac{h}{mv} \implies \lambda \propto \dfrac{1}{v}$ (mass constant).
Velocity reduced by 20%: $v' = 0.8\,v$.
$$\frac{\lambda'}{\lambda_0} = \frac{v}{v'} = \frac{v}{0.8\,v} = \frac{1}{0.8} = 1.25$$So $\lambda' = 1.25\,\lambda_0 \implies \alpha = 1.25$.
Answer: C (1.25)
Solution
Einstein’s equation gives the maximum kinetic energy $K = \dfrac{hc}{\lambda} - \phi$:
$$K_1 = \frac{hc}{\lambda_1} - \phi, \qquad K_2 = \frac{hc}{\lambda_2} - \phi$$Since $\lambda_1 = 2\lambda_2$, we have $\dfrac{hc}{\lambda_1} = \dfrac{hc}{2\lambda_2} = \dfrac{1}{2}\cdot\dfrac{hc}{\lambda_2}$.
From the second equation, $\dfrac{hc}{\lambda_2} = K_2 + \phi$, so:
$$K_1 = \frac{1}{2}(K_2 + \phi) - \phi = \frac{K_2}{2} - \frac{\phi}{2}$$Multiply by 2 and rearrange:
$$2K_1 = K_2 - \phi \implies \phi = K_2 - 2K_1$$Answer: D ($K_2 - 2K_1$)