Dual Nature of Radiation and Matter JEE Main PYQs (2020–2026)

Solved Dual Nature of Radiation and Matter PYQs

3 recent questions with answers. The full set of 159 is free to practise.

1. JEE Main 2026EasyPhoton Theory of Light and Radiation Pressure
   Q1

   A monochromatic source of light operating at 15 kW emits $2.5\times 10^{22}$ photons $/s$ . The region of an electromagnetic spectrum to which the emitted electromagnetic radiation belongs to $\_\_\_\_$ 。
   
   (Take $h=6.6\times 10^{-34}J.s$ and $c=3\times 10^{8}m/s$ ).

   • AMicrowave
   • BInfrared
   • CVisible
   • DUltraviolet

   Show answer & solution

   Correct answer: (D)

   The power $(P)$ of a light source is the total energy emitted per unit time. If the source emits n photons per second, and each photon has an energy E , then the power is:
   
   $$P=n\times E$$
   
   The given power of the source is $P=15kW=15\times 10^{3}W$ and the number of photons per second is $n=2.5\times 10^{22}{s}^{-1}$
   
   $$E=\frac{P}{n}$$
   
   $$\Rightarrow E=\frac{15\times 10^3}{2.5\times 10^{22}}$$
   
   $$\Rightarrow E=6\times 10^{-19}J$$
   
   The energy of a photon is related to its wavelength by the formula:
   
   $$E=\frac{hc}{\lambda }\Rightarrow \lambda =\frac{hc}{E}$$
   
   Substituting the constants and energy,
   
   The Plank’s constant is $h=6.6\times 10^{-34}J\cdot s$ ,
   
   The speed of light is $c=3\times 10^{8}m/s$
   
   $$\lambda =\frac{6.6\times 10^{-34}\times 3\times 10^8}{6\times 10^{-19}}m$$
   
   $$\Rightarrow \lambda =\frac{19.8\times 10^{-26}}{6\times 10^{-19}}m$$
   
   $$\Rightarrow \lambda =3.3\times 10^{-7}m$$
   
   $$\Rightarrow \lambda =3.3\times 10^{-7}\times 10^{9}nm=330nm$$
   
   Visible light typically ranges from approximately 400 nm to 700 nm . And the ultraviolet (UV) radiation ranges from approximately 10 nm to 400 nm . Since 330 nm is less than 400 nm , the radiation belongs to the ultraviolet region.
   
   Therefore, the emitted radiation belongs to the Ultraviolet region.
   
   Hence, the correct option is (D).
2. JEE Main 2026EasyDe Broglie Wavelength and Matter Waves
   Q2

   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 $\_\_\_\_$ .

   • A1.20
   • B1.0
   • C1.25
   • D0.75

   Show answer & solution

   Correct answer: (C)

   The de Broglie wavelength ( $\lambda$ ) of a particle with mass $m$ and velocity $v$ is given by:
   
   $$\lambda =\frac{h}{mv}$$
   
   Where h is Planck’s constant.
   
   For a particle of constant mass (like an electron), the wavelength is inversely proportional to its velocity:
   
   $$\lambda \propto \frac{1}{v}$$
   
   Let the initial velocity in free space be $v_0$ .
   
   The velocity in the medium ( $v_m$ ) is reduced by $20\%$ :
   
   $$v_m=v_0-20\%\text{ of }{v}_0=v_0-0.20v_0=0.80v_0$$
   
   Let ${\lambda }_0$ be the wavelength in free space and ${\lambda }_m$ be the wavelength in the medium.
   
   $$\frac{{\lambda }_m}{{\lambda }_0}=\frac{v_0}{v_m}$$
   
   Substituting the value of $v_m$ :
   
   $$\frac{{\lambda }_m}{{\lambda }_0}=\frac{v_0}{0.80v_0}=\frac{1}{0.80}$$
   
   $$\Rightarrow \frac{{\lambda }_m}{{\lambda }_0}=\frac{100}{80}=1.25$$
   
   $$\Rightarrow \lambda =1.25{\lambda }_0=\alpha {\lambda }_0$$
   
   $$\Rightarrow \alpha =1.25$$
   
   Therefore, the correct option is (C).
3. JEE Main 2026MediumEinstein’s Photoelectric Equation and Work Function
   Q3

   When a light of a given wavelength falls on a metallic surface the stopping potential for photoelectrons is 3.2 V . If a second light having wavelength twice of first light is used, the stopping potential drops to 0.7 V . The wavelength of first light is $\_\_\_\_$ m .
   
   $$\left(\begin{array}{r}h=6.63\times 10^{-34}J.s,e=1.6\times 10^{-19}C,\\ c=3\times 10^{8}m/s\end{array}\right)$$

   • A$2.9\times 10^{-8}$
   • B$2.5\times 10^{-7}$
   • C$3.1\times 10^{-7}$
   • D$2.2\times 10^{-8}$

   Show answer & solution

   Correct answer: (B)

   When a light of wavelength ( $\lambda$ ) is incident on a metal of work function $\varphi$ then the maximum kinetic energy of the photo-electron is given by Einstein’s photoelectric equation:
   
   $$K_{\max }=E-\varphi \Rightarrow E=\varphi +K_{\max }$$
   
   $$\Rightarrow \frac{hc}{\lambda }=\varphi +{eV}_s$$
   
   where $V_s$ is the stopping potential and $\varphi$ is the work function of the metal.
   
   Case 1 : Wavelength is $\lambda$ , stopping potential is 3.2 V .
   
   $\frac{hc}{\lambda }=\varphi +3.2e$ …..(i)
   
   Case 2 : Wavelength is $2\lambda$ , stopping potential is 0.7 V .
   
   $$\frac{hc}{2\lambda }=\varphi +0.7e\dots (ii)$$
   
   Subtracting equation (ii) from equation (i) :
   
   $$\left(\frac{hc}{\lambda }-\frac{hc}{2\lambda }\right)=(3.2e-0.7e)$$
   
   $$\Rightarrow \frac{hc}{2\lambda }=2.5e$$
   
   $$\Rightarrow \lambda =\frac{hc}{5e}$$
   
   Substituting the given values :
   
   $$\lambda =\frac{\left(6.63\times 10^{-34}\right)\times \left(3\times 10^8\right)}{5\times \left(1.6\times 10^{-19}\right)}$$
   
   $$\Rightarrow \lambda =\frac{19.89\times 10^{-26}}{8\times 10^{-19}}m\approx 2.5\times 10^{-7}m$$
   
   Therefore, the initial wavelength of light used was $2.5\times 10^{-7}m$ .
   
   Hence, the correct option is (B).

Topic-wise question count

Where JEE Main has asked from Dual Nature of Radiation and Matter, most frequent first.

• De Broglie Wavelength and Matter Waves63
• Einstein’s Photoelectric Equation and Work Function38
• Photoelectric Effect: Experimental Observations and Characteristics31
• Photon Theory of Light and Radiation Pressure21
• Davisson-Germer Experiment and Wave-Particle Duality6

Year-wise Dual Nature of Radiation and Matter questions in JEE Main

Number of questions from this chapter in each year's paper(s).