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 197 questions; sign in to attempt the remaining 192 in the exam simulator.

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 (04 Apr, Shift 1)Photoelectric Effect
   Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason $\mathrm{R}$ Assertion A: In photoelectric effect, on increasing the intensity of incident light the stopping potential increases. Reason R : Increase in intensity of light increases the rate of photoelectrons emitted, provided the frequency of incident light is greater than threshold frequency. In the light of the above statements, choose the correct answer from the options given below
   1. A.Both $\mathrm{A}$ and $\mathrm{R}$ are true but $\mathrm{R}$ is NOT the correct explanation of $\mathrm{A}$
   2. B.$\mathrm{A}$ is false but $\mathrm{R}$ is true
   3. C.$\mathrm{A}$ is true but $\mathrm{R}$ is false
   4. D.Both $\mathrm{A}$ and $\mathrm{R}$ are true and $\mathrm{R}$ is the correct explanation of $\mathrm{A}$
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   Answer: (B)

   $\mathrm{V}_{\mathrm{S}}=\frac{\mathrm{hv}-\phi}{\mathrm{e}}$ so stopping potential doesn't depend on Intensity $\mathrm{I}=\frac{\eta \mathrm{hv}}{\mathrm{~A}}$ On increasing intensity no. of photons per sec. $n$ increases so the no. of electrons.
3. Q3JEE Main 2026 (05 Apr, Shift 2)Matter Waves
   An electron is travelling with a velocity $v$ in free space and when it enters a medium, its velocity is reduced by $20\%$. The de Broglie wavelength of electron in the medium is $\alpha\lambda_0$, where $\lambda_0$ is its de Broglie wavelength in free space. The value of $\alpha$ is _______.
   1. A.$1.20$
   2. B.$1.0$
   3. C.$1.25$
   4. D.$0.75$
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   Answer: (C)

   The de Broglie wavelength of an electron in free space is given by $\lambda_0 = \dfrac{h}{mv}$. When the electron enters the medium, its velocity is reduced by $20\%$. The new velocity is $v' = v - 0.20v = 0.80v$. The de Broglie wavelength in the medium is $\lambda' = \dfrac{h}{mv'}$. Substituting $v' = 0.80v$, we get $\lambda' = \dfrac{h}{m(0.80v)} = \dfrac{1}{0.80} \left(\dfrac{h}{mv}\right)$. Since $\dfrac{1}{0.80} = \dfrac{100}{80} = 1.25$, we have $\lambda' = 1.25\lambda_0$. Comparing this with $\lambda' = \alpha\lambda_0$, we get $\alpha = 1.25$. Answer: $1.25$
4. Q4JEE Main 2026 (05 Apr, Shift 1)Matter Waves
   An electron of mass $m$ is moving in an electric field $\vec{E} = -2E_0\hat{i}$ ($E_0 =$ constant $\gt 0$), with an initial velocity $\vec{V} = v_0\hat{i}$ ($v_0 =$ constant $\gt 0$). If $\lambda_0 = \dfrac{h}{4mv_0}$, its de Broglie wavelength at time $t$ is __________. ($e =$ charge of electron)
   1. A.$\dfrac{4\lambda_0}{\left[1 - \dfrac{E_0 e}{2m}\dfrac{t}{v_0}\right]}$
   2. B.$\dfrac{4\lambda_0}{\left[1 + \dfrac{E_0 e}{2m}\dfrac{t}{v_0}\right]}$
   3. C.$\dfrac{4\lambda_0}{\left[1 + \dfrac{2E_0 e}{m}\dfrac{t}{v_0}\right]}$
   4. D.$\dfrac{4\lambda_0}{\left[1 - \dfrac{2E_0 e}{m}\dfrac{t}{v_0}\right]}$
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   Answer: (C)

   The force experienced by the electron in the electric field is given by $\vec{F} = q\vec{E}$. Since the charge of an electron is $-e$ and the electric field is $\vec{E} = -2E_0\hat{i}$, the force is: $\vec{F} = (-e)(-2E_0\hat{i}) = 2eE_0\hat{i}$ The acceleration of the electron is: $\vec{a} = \dfrac{\vec{F}}{m} = \dfrac{2eE_0}{m}\hat{i}$ Using the first equation of motion, the velocity of the electron at time $t$ is: $\vec{v}(t) = \vec{V} + \vec{a}t = \left(v_0 + \dfrac{2eE_0}{m}t\right)\hat{i}$ The de Broglie wavelength of the electron at time $t$ is: $\lambda = \dfrac{h}{m|\vec{v}(t)|} = \dfrac{h}{m\left(v_0 + \dfrac{2eE_0}{m}t\right)}$ We are given that $\lambda_0 = \dfrac{h}{4mv_0}$, which implies $h = 4mv_0\lambda_0$. Substituting this into the expression for $\lambda$: $\lambda = \dfrac{4mv_0\lambda_0}{m\left(v_0 + \dfrac{2eE_0}{m}t\right)}$ Dividing the numerator and the denominator by $mv_0$, we get: $\lambda = \dfrac{4\lambda_0}{1 + \dfrac{2E_0 e}{m}\dfrac{t}{v_0}}$ Answer: $\dfrac{4\lambda_0}{\left[1 + \dfrac{2E_0 e}{m}\dfrac{t}{v_0}\right]}$
5. Q5JEE Main 2026 (04 Apr, Shift 2)Matter Waves
   The de Broglie wavelength associated with an electron accelerated through a potential difference V is $\lambda_e$ and the de Broglie wavelength associated with a proton accelerated through the same potential difference is $\lambda_p$. If their corresponding masses are $m_e$ and $m_p$, respectively, then the ratio of their de Broglie wavelengths $\left(\dfrac{\lambda_e}{\lambda_p}\right)$ is ______.
   1. A.$\sqrt{\dfrac{m_p}{m_e}}$
   2. B.$\sqrt{\dfrac{m_e}{m_p}}$
   3. C.$\dfrac{m_p}{m_e}$
   4. D.$\left(\dfrac{m_p}{m_e}\right)^2$
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   Answer: (A)

   The de Broglie wavelength of a particle accelerated through a potential difference $V$ is given by $\lambda = \dfrac{h}{\sqrt{2mqV}}$. For an electron, the charge is $e$ and mass is $m_e$. Its de Broglie wavelength is $\lambda_e = \dfrac{h}{\sqrt{2m_e eV}}$. For a proton, the charge is $e$ and mass is $m_p$. Its de Broglie wavelength is $\lambda_p = \dfrac{h}{\sqrt{2m_p eV}}$. Taking the ratio of the two wavelengths, we get $\dfrac{\lambda_e}{\lambda_p} = \dfrac{\dfrac{h}{\sqrt{2m_e eV}}}{\dfrac{h}{\sqrt{2m_p eV}}} = \sqrt{\dfrac{m_p}{m_e}}$. Answer: $\sqrt{\dfrac{m_p}{m_e}}$