Dual Nature of Radiation and Matter NEET PYQs (2005–2026) with Solutions

10 solved Dual Nature of Radiation and Matter PYQs

Real NEET and AIPMT questions from this chapter, newest first. Try each one, then open the answer and the step-by-step explanation.

1. Question 1 (NEET 2026)

   NEET 2026EasyEffect of intensity on photocurrent

   A beam of light falls on a metal surface such that photo-electrons are generated. If power of the light source starts to decrease linearly with time $t$, then variation of the photocurrent $I$ and magnitude of the stopping potential $|V|$ with time is best represented by:

   1. AOption A: Graph pair A
   2. BOption B: Graph pair B
   3. COption C: Graph pair C
   4. DOption D: Graph pair D

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   Correct answer: (A) Graph pair A

   Explanation

   Each option shows an $I$–$t$ graph and a $|V|$–$t$ graph: (A) $I$ decreases linearly, $|V|$ constant; (B) both $I$ and $|V|$ decrease linearly; (C) $I$ constant, $|V|$ decreases linearly; (D) both constant. The photocurrent is proportional to the number of photons arriving per second, i.e. to the power (intensity) of the light. As the power falls linearly, $I$ falls linearly. The stopping potential depends only on the frequency: $e|V| = h\nu - \phi_0$. The frequency does not change, so $|V|$ stays constant. Hence option (A).

2. Question 2 (NEET 2026)

   NEET 2026EasyWave nature of light, particle nature of light and de Broglie wavelength

   Match List I with List II. Choose the correct answer from the options given below.Match List I with List II. Choose the correct answer from the options given below.

   1. AOption A: A-IV, B-I, C-II, D-III
   2. BOption B: A-IV, B-III, C-II, D-I
   3. COption C: A-I, B-IV, C-III, D-II
   4. DOption D: A-IV, B-III, C-I, D-II

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   Correct answer: (D) A-IV, B-III, C-I, D-II

   Explanation

   ["A. E = hν → Energy of photon → IV","B. Diffraction and interference confirm wave nature of light → III","C. λ = h/p represents de Broglie wavelength → I","D. Compton effect confirms particle nature of light → II","","Correct matching:"]

3. Question 3 (NEET 2026)

   NEET 2026MediumThreshold wavelength and stopping potential

   A ray of light with wavelength $\lambda$ is incident on three different photoelectric cells namely 1, 2 and 3. The threshold wavelength of these photoelectric cells are $\lambda_1$, $\lambda_2$, and $\lambda_3$, respectively and the magnitude of stopping potentials of these cells are $V_1$, $V_2$ and $V_3$, respectively. The relation between $\lambda$ and threshold wavelengths are $\lambda_1 < \lambda$, $\lambda_2 > \lambda$ and $\lambda_3 \gg \lambda$. The correct option is:

   1. AOption A: $V_1 = 0, V_2 < V_3$
   2. BOption B: $V_1 = 0, V_2 > V_3$
   3. COption C: $V_1 > V_2, V_3 = 0$
   4. DOption D: $V_1 < V_2, V_3 = 0$

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   Correct answer: (A) $V_1 = 0, V_2 < V_3$

   Explanation

   Photoemission happens only if the incident wavelength is shorter than the threshold wavelength. Cell 1 has $\lambda_1 < \lambda$, so no photoelectrons are emitted: $V_1 = 0$. For cells 2 and 3: $eV = hc\left(\frac{1}{\lambda} - \frac{1}{\lambda_0}\right)$. A larger threshold wavelength (smaller work function) gives a larger stopping potential. Since $\lambda_3 \gg \lambda$ while $\lambda_2$ is only greater than $\lambda$, $V_3 > V_2$, i.e. $V_2 < V_3$. Hence option (A).

4. Question 4 (NEET 2026)

   NEET 2026MediumMomentum of photon and electron

   A photon and an electron, each of $20\ \mathrm{eV}$ energy, move in free space. The ratio of linear momentum of electron $p_e$ to that of photon $p_{Ph}$, $\frac{p_e}{p_{Ph}}$ is: [Take speed of light = $3 \times 10^8\ \mathrm{ms^{-1}}$, charge of electron = $-1.6 \times 10^{-19}\ \mathrm{C}$ and mass of electron = $9 \times 10^{-31}\ \mathrm{kg}$]

   1. AOption A: $\frac{2}{450}$
   2. BOption B: $\frac{1}{250}$
   3. COption C: $225$
   4. DOption D: $275$

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   Correct answer: (C) $225$

   Explanation

   $E = 20\ \mathrm{eV} = 20 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-18}\ \mathrm{J}$. Photon: $p_{Ph} = \frac{E}{c}$. Electron (non-relativistic): $p_e = \sqrt{2mE}$. $$\frac{p_e}{p_{Ph}} = c\sqrt{\frac{2m}{E}} = 3 \times 10^8 \times \sqrt{\frac{2 \times 9 \times 10^{-31}}{3.2 \times 10^{-18}}} = 3 \times 10^8 \times \sqrt{5.625 \times 10^{-13}}$$ $\sqrt{5.625 \times 10^{-13}} = 7.5 \times 10^{-7}$, so$\frac{p_e}{p_{Ph}} = 3 \times 10^8 \times 7.5 \times 10^{-7} = 225$. (Check:$p_e = 2.4 \times 10^{-24}$and$p_{Ph} \approx 1.07 \times 10^{-26}\ \mathrm{kg\,m\,s^{-1}}$.) Option (C).

5. Question 5 (NEET 2026)

   NEET 2026MediumPhotoelectric Effect

   For a metal of work function 6.6 eV, which of the following wavelengths of incident radiation does not give rise to the photoelectric effect? (Take Planck's constant as 6.6 × 10⁻³⁴ J s)

   1. AOption A: 100 nm
   2. BOption B: 150 nm
   3. COption C: 200 nm
   4. DOption D: 50 nm

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   Correct answer: (C) 200 nm

   Explanation

   Threshold wavelength λ₀=hc/W₀=(6.6×10⁻³⁴×3×10⁸)/(6.6×1.6×10⁻¹⁹)≈187.5 nm. Photoelectric effect occurs only for λ<187.5 nm. Therefore 200 nm does not produce photoelectric emission. Hence option (C) is correct.

6. Question 6 (NEET 2026)

   NEET 2026EasyPhoton Theory

   A bulb is rated at 150 watt, converting 8% energy into light. If energy of one photon is 4.42 × 10⁻¹⁹ J, how many photons are emitted by the bulb per second?

   1. AOption A: 2.71 × 10¹⁹
   2. BOption B: 4.06 × 10¹⁹
   3. COption C: 27.2 × 10¹⁹
   4. DOption D: 1.35 × 10¹⁹

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   Correct answer: (A) 2.71 × 10¹⁹

   Explanation

   Energy converted to light per second = 150×0.08=12 J. Number of photons = 12/(4.42×10⁻¹⁹)=2.71×10¹⁹ photons/s. Hence option (A) is correct.

7. Question 7 (NEET 2025)

   NEET 2025MediumMatter Waves

   De-Broglie wavelength of an electron orbiting in the n = 2 state of hydrogen atom is close to (Given Bohr radius = 0.052 nm)

   1. AOption A: 0.067 nm
   2. BOption B: 0.67 nm
   3. COption C: 1.67 nm
   4. DOption D: 2.67 nm

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   Correct answer: (B) 0.67 nm

   Explanation

   For an electron in the nth Bohr orbit, the de Broglie wavelength is given by: λ = 2πr/n For n = 2, radius r = n²a₀ = 4 × 0.052 nm = 0.208 nm Hence, λ = (2π × 0.208)/2 ≈ 0.653 nm ≈ 0.67 nm Therefore, option (B) is correct.

8. Question 8 (NEET 2025)

   NEET 2025MediumMatter Waves

   A photon and an electron (mass $m$) have the same energy $E$. The ratio $\left(\lambda_{\text{photon}}/\lambda_{\text{electron}}\right)$ of their de Broglie wavelengths is: ($c$ is the speed of light)

   1. AOption A: $\sqrt{\frac{E}{2m}}$
   2. BOption B: $c\sqrt{2mE}$
   3. COption C: $c\sqrt{\frac{2m}{E}}$
   4. DOption D: $\frac{1}{c}\sqrt{\frac{E}{2m}}$

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   Correct answer: (C) $c\sqrt{\frac{2m}{E}}$

   Explanation

   For a photon: $$E = pc$$ Hence momentum of photon: $$p_{\text{photon}} = \frac{E}{c}$$ Therefore, $$\lambda_{\text{photon}} = \frac{h}{p_{\text{photon}}} = \frac{hc}{E}$$ For an electron (non-relativistic): $$E = \frac{p^2}{2m}$$ So, $$p_{\text{electron}} = \sqrt{2mE}$$ Hence, $$\lambda_{\text{electron}} = \frac{h}{\sqrt{2mE}}$$ Therefore, $$\frac{\lambda_{\text{photon}}}{\lambda_{\text{electron}}} = \frac{hc/E}{h/\sqrt{2mE}}$$ $$= c\sqrt{\frac{2m}{E}}$$ Thus, option (C) is correct.

9. Question 9 (NEET 2023)

   NEET 2023EasyX-rays

   The minimum wavelength of X-rays produced by an electron accelerated through a potential difference of V volts is proportional to:

   1. AOption A: √V
   2. BOption B: 1/V
   3. COption C: 1/√V
   4. DOption D: V²

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   Correct answer: (B) 1/V

   Explanation

   For X-rays, the minimum wavelength is given by: λmin = hc/eV Therefore, λmin ∝ 1/V Hence, option (B) is correct.

10. Question 10 (NEET 2022)

    NEET 2022MediumEinstein's Photoelectric Equation

    When two monochromatic lights of frequency $\nu$ and $\frac{\nu}{2}$ are incident on a photoelectric metal, their stopping potentials become $\frac{V_s}{2}$ and $V_s$ respectively. The threshold frequency for this metal is:

    1. AOption A: $\frac{3}{2}\nu$
    2. BOption B: $2\nu$
    3. COption C: $3\nu$
    4. DOption D: $\frac{2}{3}\nu$

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    Correct answer: (A) $\frac{3}{2}\nu$

    Explanation

    Using Einstein’s photoelectric equation: $eV = h(\nu-\nu_0)$ For frequency $\nu$: $e\left(\frac{V_s}{2}\right)=h(\nu-\nu_0)$ ...(1) For frequency $\frac{\nu}{2}$: $eV_s=h\left(\frac{\nu}{2}-\nu_0\right)$ ...(2) Multiplying (1) by 2: $eV_s=2h(\nu-\nu_0)$ Equating with (2): $2(\nu-\nu_0)=\frac{\nu}{2}-\nu_0$ $2\nu-2\nu_0=\frac{\nu}{2}-\nu_0$ $\frac{3\nu}{2}=\nu_0$ Hence threshold frequency: $\boxed{\nu_0=\frac{3}{2}\nu}$ Therefore option (A) is correct.