JEE Previous Year Questions

How JEE Tests This Chapter

Dual Nature contributes 1-2 questions to nearly every JEE Main paper — usually numericals built on a small set of templates:

1. Einstein-equation plug-ins: $K_{max} = \dfrac{hc}{\lambda} - \phi_0$, often with the 1240 eV nm shortcut and answers demanded in eV or as stopping potentials.
2. Two-wavelength/two-reading eliminations: subtract two applications of the equation to kill $\phi_0$ and extract h, $\phi_0$ or a ratio.
3. Graph readings: slope $h/e$, intercepts $\nu_0$ and $-\phi_0/e$; parallel lines for different metals.
4. Photon counting and momentum: $N = P/h\nu$, radiation force = momentum delivered per second ($P/c$ absorbed, $2P/c$ reflected).
5. de Broglie scaling: $\lambda = \dfrac{h}{\sqrt{2mK}} = \dfrac{h}{\sqrt{2mqV}}$, ratios between electron/proton/alpha/deuteron at equal V, K, p or v; the $\lambda \propto 1/\sqrt{V}$ shifts.
6. Percentage/factor changes: 'wavelength decreased by x%, stopping potential changes by…?'

All questions below are JEE-style previous-year questions. Exam tags identify the pattern; years are attached only where attribution is certain — otherwise the tag is the generic [JEE Mains] / [JEE Advanced].

Constants used: h = $6.63 \times 10^{-34}$ J s, hc ≈ 1240 eV nm, $m_e = 9.1 \times 10^{-31}$ kg, e = $1.6 \times 10^{-19}$ C.

JEE PYQ Worked Set A: Photoelectric Core

PYQ 1. The work function of a metal is 4.0 eV. The longest wavelength that can cause photoemission is: [JEE Mains]

Solution:

1. $\lambda_0 = \dfrac{hc}{\phi_0} = \dfrac{1240}{4.0} = 310$ nm.
2. Any $\lambda > 310$ nm fails regardless of intensity.

PYQ 2. Light of wavelength 300 nm falls on a metal of work function 2.13 eV. The maximum kinetic energy and stopping potential are: [JEE Mains]

Solution:

1. Photon energy $= \dfrac{1240}{300} = 4.13$ eV.
2. $K_{max} = 4.13 - 2.13 = 2.0$ eV; $V_0 = 2.0$ V.

PYQ 3. When light of wavelength $\lambda$ illuminates a metal, the stopping potential is $V$. When the wavelength becomes $\lambda/2$, the stopping potential becomes $3V$. The threshold wavelength is: [JEE Mains]

Solution:

1. $eV = \dfrac{hc}{\lambda} - \phi_0$ … (i); $3eV = \dfrac{2hc}{\lambda} - \phi_0$ … (ii).
2. (ii) - 3×(i): $0 = \dfrac{2hc}{\lambda} - \dfrac{3hc}{\lambda} + 2\phi_0 \Rightarrow \phi_0 = \dfrac{hc}{2\lambda}$.
3. Threshold: $\lambda_0 = \dfrac{hc}{\phi_0} = 2\lambda$.
4. Takeaway: simultaneous-equation elimination of $\phi_0$ — the most re-used JEE template in this chapter.

PYQ 4. The stopping potentials are $V_1$ and $V_2$ for incident frequencies $\nu_1$ and $\nu_2$. Planck's constant equals: [JEE Mains]

Solution:

1. $eV_1 = h\nu_1 - \phi_0$ and $eV_2 = h\nu_2 - \phi_0$.
2. Subtract: $e(V_1 - V_2) = h(\nu_1 - \nu_2)$.
3. $h = \dfrac{e(V_1 - V_2)}{\nu_1 - \nu_2}$ — Millikan's slope method in symbols.

PYQ 5. A 200 W sodium lamp radiates uniformly in all directions at 589 nm. The photon flux (photons per m² per s) at 2 m from the lamp is closest to: [JEE Mains]

Solution:

1. Photon energy $E = \dfrac{1240}{589} \approx 2.1$ eV $= 3.37 \times 10^{-19}$ J.
2. Total rate $N = \dfrac{200}{3.37 \times 10^{-19}} = 5.9 \times 10^{20}$ /s.
3. Flux at r = 2 m: $\dfrac{N}{4\pi r^2} = \dfrac{5.9 \times 10^{20}}{4\pi \times 4} \approx 1.2 \times 10^{19}$ photons m⁻² s⁻¹.
4. Takeaway: photon counting + inverse-square spreading — two standard steps chained.

PYQ 6. A perfectly reflecting mirror receives a normally incident laser beam of power P. The radiation force on the mirror is: [JEE Mains]

Solution:

1. Momentum arrives at rate $P/c$; reflection reverses it, delivering $2P/c$.
2. Force $F = \dfrac{2P}{c}$. For absorption it would be $P/c$.
3. Takeaway: photon momentum p = E/c scaled up to beam level — remember the factor 2 for mirrors.

JEE PYQ Worked Set B: de Broglie Patterns

PYQ 7. An electron and a photon each have wavelength 1.0 nm. The ratio of photon energy to electron kinetic energy is closest to: [JEE Mains]

Solution:

1. Photon: $E_{ph} = \dfrac{1240}{1.0 \text{ nm}} = 1240$ eV.
2. Electron: from $\lambda = \dfrac{1.227}{\sqrt{V}}$ nm, $\lambda = 1$ nm → $V = 1.5$ V → $K_e \approx 1.5$ eV.
3. Ratio: $\dfrac{1240}{1.5} \approx 820$.
4. Takeaway: at equal wavelength the photon is vastly more energetic — E scales as $1/\lambda$ for photons but $1/\lambda^2$ for slow electrons.

PYQ 8. The de Broglie wavelength of an electron accelerated through V volts is 0.1227 nm. V equals: [JEE Mains]

Solution:

1. $V = \left(\dfrac{1.227}{0.1227}\right)^2 = 10^2 = 100$ V.

PYQ 9. A proton and a deuteron (mass 2$m_p$, charge e) are accelerated through the same potential. The ratio $\lambda_d/\lambda_p$ is: [JEE Mains]

Solution:

1. $\lambda \propto \dfrac{1}{\sqrt{mq}}$ at fixed V; charges are equal.
2. $\dfrac{\lambda_d}{\lambda_p} = \sqrt{\dfrac{m_p}{2m_p}} = \dfrac{1}{\sqrt{2}}$.
3. Takeaway: the deuteron's wavelength is shorter by $\sqrt{2}$ — mass under a square root, charge equal so it drops out.

PYQ 10. If the kinetic energy of a moving particle is increased by 300%, its de Broglie wavelength changes by: [JEE Mains]

Solution:

1. K → 4K (a 300% increase means quadrupling).
2. $\lambda \propto \dfrac{1}{\sqrt{K}}$ → $\lambda' = \dfrac{\lambda}{2}$.
3. The wavelength decreases by 50%.
4. Takeaway: translate percentage language into factors before touching the formula — '300% increase' trips more students than the physics does.

PYQ 11. An electron (mass m) and a photon have the same energy E. The ratio of the electron's de Broglie wavelength to the photon's wavelength is: [JEE Mains]

Solution:

1. Electron: $\lambda_e = \dfrac{h}{\sqrt{2mE}}$.
2. Photon: $\lambda_{ph} = \dfrac{hc}{E}$.
3. Ratio: $\dfrac{\lambda_e}{\lambda_{ph}} = \dfrac{h}{\sqrt{2mE}} \cdot \dfrac{E}{hc} = \dfrac{1}{c}\sqrt{\dfrac{E}{2m}}$.
4. Takeaway: a pure symbol-manipulation classic — write both wavelengths, divide, simplify.

PYQ 12. The ratio of the de Broglie wavelengths of an electron and a proton, each moving with the same kinetic energy, is (take $m_p = 1836\,m_e$): [JEE Mains]

Solution:

1. $\dfrac{\lambda_e}{\lambda_p} = \sqrt{\dfrac{m_p}{m_e}} = \sqrt{1836} \approx 42.8$.
2. About 43 : 1 in the electron's favour.

JEE PYQ Worked Set C: Advanced-flavour Multi-step

PYQ 13. Light of wavelength 330 nm on metal A ($\phi_0$ = 2.5 eV) and 220 nm on metal B ($\phi_0$ = 3.5 eV): the ratio of maximum speeds of photoelectrons $v_A : v_B$ is: [JEE Advanced pattern]

Solution:

1. $K_A = \dfrac{1240}{330} - 2.5 = 3.76 - 2.5 = 1.26$ eV.
2. $K_B = \dfrac{1240}{220} - 3.5 = 5.64 - 3.5 = 2.14$ eV.
3. $\dfrac{v_A}{v_B} = \sqrt{\dfrac{K_A}{K_B}} = \sqrt{\dfrac{1.26}{2.14}} \approx 0.77$, i.e. roughly 3 : 4.
4. Takeaway: speeds compare under a square root of energies — never linearly.

PYQ 14. In a photoelectric experiment, reducing the wavelength from 500 nm to 400 nm increases the stopping potential by: [JEE Mains]

Solution:

1. $e\Delta V_0 = hc\left(\dfrac{1}{400} - \dfrac{1}{500}\right)$ (nm units, eV output via 1240).
2. $\Delta V_0 = 1240 \times \left(\dfrac{1}{400} - \dfrac{1}{500}\right) = 1240 \times \dfrac{100}{200000} = 0.62$ V.
3. Takeaway: the work function cancels in differences — a metal-independent answer.

PYQ 15. A photosensitive surface is illuminated in turn by light of wavelengths $\lambda$ and $\lambda/2$. If the maximum kinetic energies are in ratio 1 : 3, the threshold wavelength is: [JEE Mains]

Solution:

1. $K_1 = \dfrac{hc}{\lambda} - \phi_0$; $K_2 = \dfrac{2hc}{\lambda} - \phi_0 = 3K_1$.
2. Substitute: $\dfrac{2hc}{\lambda} - \phi_0 = \dfrac{3hc}{\lambda} - 3\phi_0 \Rightarrow 2\phi_0 = \dfrac{hc}{\lambda} \Rightarrow \phi_0 = \dfrac{hc}{2\lambda}$.
3. $\lambda_0 = \dfrac{hc}{\phi_0} = 2\lambda$.
4. Takeaway: identical machinery to PYQ 3 — recognise the template, not the numbers.

PYQ 16. An X-ray tube operates at 50 kV. The minimum wavelength of the emitted X-rays is: [JEE Mains]

Solution:

1. $\lambda_{min} = \dfrac{1240 \text{ eV nm}}{50000 \text{ eV}} = 0.0248$ nm.
2. ≈ 0.025 nm — the Duane-Hunt limit, i.e. the inverse photoelectric effect.