Dual Nature of Matter JEE Main previous year questions with solutions

5 solved JEE Main questions on Dual Nature of Matter, free to read — no sign-in needed. The full chapter has 207 questions; sign in to attempt the remaining 202 in the exam simulator.

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1. Q1JEE Main 2026 (06 Apr, Shift 2)Matter Waves
   The de Broglie wavelength for an electron accelerated through the potential difference of $V_1$ volt is $\lambda_1$. When the potential difference is changed to $V_2$ volt, the associated de Broglie wavelength is increased by $50\%$. If $(V_1/V_2) = (9/\alpha)$, then the value of $\alpha$ is __________.
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   Answer: 4

   The de Broglie wavelength of an electron accelerated through a potential difference $V$ is given by $\lambda = \dfrac{h}{\sqrt{2meV}}$. This implies $\lambda \propto \dfrac{1}{\sqrt{V}}$. Given that the new wavelength $\lambda_2$ is increased by $50\%$, we have $\lambda_2 = \lambda_1 + 0.5\lambda_1 = \dfrac{3}{2}\lambda_1$. Taking the ratio of the wavelengths, we get $\dfrac{\lambda_1}{\lambda_2} = \sqrt{\dfrac{V_2}{V_1}}$. Substituting $\lambda_2 = \dfrac{3}{2}\lambda_1$, we get $\dfrac{2}{3} = \sqrt{\dfrac{V_2}{V_1}}$. Squaring both sides yields $\dfrac{V_2}{V_1} = \dfrac{4}{9}$, which gives $\dfrac{V_1}{V_2} = \dfrac{9}{4}$. Comparing this with the given relation $\dfrac{V_1}{V_2} = \dfrac{9}{\alpha}$, we find $\alpha = 4$. Answer: $4$
2. Q2JEE Main 2025 (29 Jan, Shift 1)Atomic Models
   Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Emission of electrons in photoelectric effect can be suppressed by applying a sufficiently negative electron potential to the photoemissive substance. Reason (R) : A negative electric potential, which stops the emission of electrons from the surface of a photoemissive substance, varies linearly with frequency of incident radiation. In the light of the above statements, choose the most appropriate answer from the options given below :
   1. A.(A) is false but (R) is true
   2. B.(A) is true but (R) is false
   3. C.Both $(\mathbf{A})$ and $(\mathbf{R})$ are true and $(\mathbf{R})$ is the correct explanation of $(\mathbf{A})$
   4. D.Both $(\mathbf{A})$ and $(\mathbf{R})$ are true but $(\mathbf{R})$ is not the correct explanation of $(\mathbf{A})$
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   Answer: (D)

   Assertion (A) is true because a sufficiently negative potential can repel and stop the emitted electrons from leaving the surface. Reason (R) is also true because the stopping potential $V_{\text {stop }}$ varies linearly with the frequency $\nu$ of the incident light $\left(V_{\text {stop }}=\frac{h}{e} \nu-\frac{\phi}{e}\right)$. However, (R) does not directly explain why a negative potential suppresses electron emission; it only shows how much potential is needed for different frequencies. Therefore, both (A) and (R) are true, but (R) is not the correct explanation of (A).
3. Q3JEE Main 2024 (09 Apr, Shift 2)Photoelectric Effect
   UV light of $4.13 \mathrm{eV}$ is incident on a photosensitive metal surface having work function $3.13 \mathrm{eV}$. The maximum kinetic energy of ejected photoelectrons will be:
   1. A.$4.13 \mathrm{eV}$
   2. B.$3.13 \mathrm{eV}$
   3. C.$1 \mathrm{eV}$
   4. D.$7.26 \mathrm{eV}$
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   Answer: (C)

   $\begin{aligned} & \mathrm{E}_{\text {photon }}=(\text { work function })+\mathrm{K} \cdot \mathrm{E}_{\max } \\ & \therefore 4.13=3.13+\mathrm{K} \cdot \mathrm{E}_{\max } \\ & \therefore \mathrm{K}_{\mathrm{max}}=1 \mathrm{eV}\end{aligned}$
4. Q4JEE Main 2023 (31 Jan, Shift 1)Diffraction of Light
   Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R Assertion A : The beam of electrons shows wave nature and exhibit interference and diffraction. Reason R : Davisson Germer Experimentally verified the wave nature of electrons. In the light of the above statements. Choose the most appropriate answer from the options given below :
   1. A.A is correct but R is not correct
   2. B.A is not correct but R is correct
   3. C.Both A and R are correct but R is Not the correct explanation of A
   4. D.Both A and R are correct and R is the correct explanation of A
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   Answer: (D)

   The Davisson-Germer experiment showed that electron beams can undergo diffraction when passed through the atomic crystals. This shows that the wave nature of electrons. When electron is in wave nature it can exhibit interference and show diffraction. Hence, Both A and R are correct and R is the correct explanation of A.
5. Q5JEE Main 2022 (26 Jun, Shift 2)Motion of Charged Particle in Magnetic Field
   A metal surface is illuminated by a radiation of wavelength $4500A^{^{\circ}}$. The ejected photo-electron enters a constant magnetic field of $2mT$ making an angle of $90^{\circ}$ with the magnetic field. If it starts revolving in a circular path of radius $2mm$, the work function of the metal is approximately
   1. A.$1.36eV$
   2. B.$1.69eV$
   3. C.$2.78eV$
   4. D.$2.23eV$
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   Answer: (A)

   From the Einstein's photoelectric equation $K=\dfrac{hc}{\lambda }-\phi \Rightarrow \phi =\dfrac{hc}{\lambda }-K$ Now $R=\dfrac{mv}{Bq}\Rightarrow v=\dfrac{RBq}{m}\Rightarrow K=\dfrac{{R}^{2}{B}^{2}{q}^{2}}{2m}$ Therefore, $\phi =\dfrac{hc}{\lambda }-\dfrac{{R}^{2}{B}^{2}{q}^{2}}{2m}=\dfrac{6.63\times {10}^{-34}\times 3\times {10}^{8}}{4500\times {10}^{-10}}-\dfrac{{\left(2\times {10}^{-3}\times 2\times {10}^{-3}\times 1.6\times {10}^{-19}\right)}^{2}}{2\times 9.1\times {10}^{-31}} \Rightarrow \phi =\left(4.42-2.25\right)\times {10}^{-19}J=\dfrac{2.27}{1.6}eV=1.36eV$