NEET Previous Year Questions

How NEET Tests This Chapter

Dual Nature delivers 1-2 questions in nearly every NEET paper, drawn from a tight, predictable pool:

1. Photon plug-ins: $E = h\nu = hc/\lambda$, $p = h/\lambda$, photons per second from power.
2. Einstein's equation: $K_{max} = h\nu - \phi_0$, stopping potential reads, threshold conditions.
3. The distinctions: what intensity changes vs what frequency changes; what happens below threshold.
4. Graphs: $V_0$ vs $\nu$ slope and intercepts; I-V curve features.
5. de Broglie comparisons: electron vs proton vs alpha at equal speed / energy / momentum / accelerating potential; scaling with V and K.
6. Fact recall: photon properties (neutral, rest mass zero, momentum $h\nu/c$), Davisson-Germer key numbers, Nobel history.

Everything below is NEET-style previous-year material with fully worked explanations. Years are attached only where attribution is certain; otherwise the tag is the generic [NEET].

Constants: 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.

NEET PYQ Worked Set A: Photon & Einstein Equation

PYQ 1. The energy of a photon of wavelength 663 nm is: [NEET]

Solution:

1. $E = \dfrac{hc}{\lambda} = \dfrac{(6.63 \times 10^{-34})(3 \times 10^8)}{663 \times 10^{-9}} = 3.0 \times 10^{-19}$ J ≈ 1.87 eV.
2. The 663 nm choice makes the SI arithmetic cancel neatly — a NEET signature.

PYQ 2. The threshold frequency of a metal is $6.2 \times 10^{14}$ Hz. Its work function is about: [NEET]

Solution:

1. $\phi_0 = h\nu_0 = (6.63 \times 10^{-34})(6.2 \times 10^{14}) = 4.11 \times 10^{-19}$ J.
2. In eV: $\dfrac{4.11 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.6$ eV.

PYQ 3. Light of frequency 1.5 times the threshold frequency falls on a photosensitive material. If the frequency is halved and intensity doubled, the photocurrent becomes: [NEET]

Solution:

1. New frequency $= \dfrac{1.5\nu_0}{2} = 0.75\nu_0 < \nu_0$ — below threshold.
2. Below threshold, intensity is irrelevant: photocurrent = zero.
3. Takeaway: NEET's favourite trap — always check the threshold before computing anything.

PYQ 4. The photoelectric work function of a metal is 3.3 eV. The threshold wavelength is closest to: [NEET]

Solution:

1. $\lambda_0 = \dfrac{1240}{3.3} \approx 376$ nm — ultraviolet edge.

PYQ 5. When light of wavelength 400 nm illuminates a metal of work function 2.1 eV, the stopping potential is: [NEET]

Solution:

1. Photon energy $= \dfrac{1240}{400} = 3.1$ eV.
2. $K_{max} = 3.1 - 2.1 = 1.0$ eV → $V_0 = 1.0$ V.

PYQ 6. A 100 W light source emits photons of average energy $3.3 \times 10^{-19}$ J. The photon emission rate is: [NEET]

Solution:

1. $N = \dfrac{P}{E} = \dfrac{100}{3.3 \times 10^{-19}} \approx 3 \times 10^{20}$ photons per second.

PYQ 7. The momentum of a photon of energy 1 MeV, in kg m/s, is: [NEET]

Solution:

1. $p = \dfrac{E}{c} = \dfrac{10^6 \times 1.6 \times 10^{-19}}{3 \times 10^8}$.
2. $p = 5.33 \times 10^{-22}$ kg m/s.
3. Takeaway: for photons, always p = E/c — never $\sqrt{2mE}$ (no mass to use!).

NEET PYQ Worked Set B: de Broglie Patterns

PYQ 8. An electron of mass m and a photon have the same energy E. The ratio of the de Broglie wavelength of the electron to the wavelength of the photon is (c = speed of light): [NEET 2019 pattern]

Solution:

1. Electron: $\lambda_e = \dfrac{h}{\sqrt{2mE}}$; Photon: $\lambda_{ph} = \dfrac{hc}{E}$.
2. Ratio: $\dfrac{\lambda_e}{\lambda_{ph}} = \dfrac{1}{c}\left(\dfrac{E}{2m}\right)^{1/2}$.

PYQ 9. An electron is accelerated through a potential difference of 10,000 V. Its de Broglie wavelength is about: [NEET]

Solution:

1. $\lambda = \dfrac{1.227}{\sqrt{10^4}} = \dfrac{1.227}{100}$ nm.
2. $\lambda \approx 0.0123$ nm = $12.3 \times 10^{-12}$ m — X-ray scale.

PYQ 10. Which particle, all moving with the same velocity, has the longest de Broglie wavelength: electron, proton, deuteron or alpha? [NEET]

Solution:

1. $\lambda \propto \dfrac{1}{m}$ at equal velocity.
2. Lightest particle wins: the electron.

PYQ 11. A proton and an alpha particle have equal kinetic energy. The ratio $\lambda_p : \lambda_\alpha$ is: [NEET]

Solution:

1. $\lambda \propto \dfrac{1}{\sqrt{m}}$ at equal K.
2. $\dfrac{\lambda_p}{\lambda_\alpha} = \sqrt{\dfrac{m_\alpha}{m_p}} = \sqrt{4} = 2$ → ratio 2 : 1.

NEET PYQ Worked Set C: Concept Discriminators

PYQ 12. In a photoelectric experiment, the collector potential is made increasingly negative. The photocurrent: [NEET]

Solution:

1. Only electrons with kinetic energy above eV (V = retarding potential) reach the collector.
2. The current decreases rapidly and becomes zero at the sharply defined stopping potential $V_0$, where $eV_0 = K_{max}$.

PYQ 13. The photoelectric effect can be explained on the basis of: [NEET]

Solution:

1. The quantum (corpuscular/photon) theory of light — energy absorbed as whole quanta $h\nu$ by single electrons.
2. The wave theory fails on intensity-independence of $K_{max}$, the threshold, and instantaneity.

PYQ 14. When ultraviolet light falls on a caesium surface, photoelectrons are emitted. To INCREASE their maximum kinetic energy one should: [NEET]

Solution:

1. $K_{max} = h\nu - \phi_0$ — only frequency (or the material) moves it.
2. Increase the frequency (decrease the wavelength) of the light; increasing intensity only adds more electrons at the same energies.

PYQ 15. The Davisson-Germer experiment (nickel crystal, 54 V, peak at 50 degrees) established: [NEET; beyond rationalised Board syllabus]

Solution:

1. Electrons scattered from the crystal formed a diffraction maximum — wave behaviour.
2. Measured wavelength 0.165 nm ≈ de Broglie's $\dfrac{1.227}{\sqrt{54}} = 0.167$ nm.
3. Conclusion: experimental confirmation of the wave nature of electrons (matter waves).

PYQ 16. A photocell's saturation current with a source at 0.6 m is 12 mA. At 1.2 m it becomes: [NEET pattern]

Solution:

1. Intensity ∝ $1/r^2$: doubling distance quarters intensity.
2. Saturation current ∝ intensity → $\dfrac{12}{4} = 3$ mA.