Dual Nature of Radiation and Matter Previous Year Questions — JEE Main Physics
31 past-year Dual Nature of Radiation and Matter questions from JEE Main (Physics) — page 3.
Q17 (2025)
The energy E and momentum p of a moving body of mass m are related by some equation. Given that c represents the speed of light, identify the correct equation.
- $E^{2} = pc^{2} + m^{2}c^{4}$
- $E^{2} = pc^{2} + m^{2}c^{2}$
- $E^{2} = p^{2}c^{2} + m^{2}c^{2}$
- $E^{2} = p^{2}c^{2} + m^{2}c^{4}$
### Related Formula
Relativistic total energy equation:
$$E^2 = p^2c^2 + m_0^2c^4$$
### Core Logic
We can verify this equation via dimensional analysis:
- Dimension of energy, $[E] = M^1 L^2 T^{-2} \implies [E^2] = M^2 L^4 T^{-4}$
- Dimension of momentum times light speed, $[pc] = (M^1 L^1 T^{-1}) \cdot (L^1 T^{-1}) = M^1 L^2 T^{-2} \implies [p^2c^2] = M^2 L^4 T^{-4}$
- Rest energy square term, $[m^2c^4] = M^2 \cdot (L^1 T^{-1})^4 = M^2 L^4 T^{-4}$
Since all terms share identical dimensions, this relativistic identity is structurally valid. The matching option is (4).
### Pattern Recognition
This is Einstein's famous energy-momentum relation of special relativity, fundamental to high-energy modern physics contexts.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Dual Nature of Radiation and Matter
Q23 (2025)
The ratio of the power of a light source $S_{1}$ to that the light source $S_{2}$ is 2. $S_{1}$ is emitting $2 \times 10^{15}$ photons per second at 600 nm. If the wavelength of the source $S_{2}$ is 300 nm, then the number of photons per second emitted by $S_{2}$ is ____ $\times 10^{14}$ .
### Related Formula
Power of a light source:
$$P = n E_{\text{photon}} = n \left(\frac{hc}{\lambda}\right)$$
where $n$ is the number of photons emitted per second.
### Core Logic
Taking the ratio of power for source 1 and source 2:
$$\frac{P_1}{P_2} = \frac{n_1 \left(\frac{hc}{\lambda_1}\right)}{n_2 \left(\frac{hc}{\lambda_2}\right)} = \left(\frac{\lambda_2}{\lambda_1}\right) \frac{n_1}{n_2}$$
Given data:
- $\frac{P_1}{P_2} = 2$
- $n_1 = 2 \times 10^{15}\ \mathrm{photons/s}$
- $\lambda_1 = 600\ \mathrm{nm}$
- $\lambda_2 = 300\ \mathrm{nm}$
Substitute the values:
$$2 = \left(\frac{300}{600}\right) \times \frac{2 \times 10^{15}}{n_2}$$
$$2 = \frac{1}{2} \times \frac{2 \times 10^{15}}{n_2} \implies n_2 = \frac{10^{15}}{2} = 5 \times 10^{14}\ \mathrm{photons/s}$$
The question asks for the value multiplying $10^{14}$, which is 5.
### Pattern Recognition
Power relies on both the delivery rate and individual photon packet energy. Lower wavelength means more energetic packets, requiring fewer photons to emit the same power.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Dual Nature of Radiation and Matter
Q1142 (2025)
An electron of mass 'm' with an initial velocity $\vec{v}=v_{0}\hat{i}(v_{0}>0)$ enters an electric field $\vec{E}=-E_{0}\hat{k}$ If the initial de Broglie wavelength is $\lambda_{0}$, the value after time t would be :-
- $\frac{\lambda_{0}}{\sqrt{1+\frac{e^{2}E_{0}^{2}t^{2}}{m^{2}v_{0}^{2}}}}$
- $\frac{\lambda_{0}}{\sqrt{1-\frac{e^{2}E_{0}^{2}t^{2}}{m^{2}v_{0}^{2}}}}$
- $\lambda_{0}$
- $\lambda_{0}\sqrt{1+\frac{e^{2}E_{0}^{2}t^{2}}{m^{2}v_{0}^{2}}}$
### Related Formula
The de Broglie wavelength relation matching a moving particle momentum is given by:
$$\lambda = \frac{h}{p} = \frac{h}{m|\vec{v}|}$$
### Core Logic
Calculate the electric acceleration force acting component on the charge :
$$\vec{a} = \frac{q\vec{E}}{m} = \frac{(-e)(-E_{0}\hat{k})}{m} = \frac{eE_{0}}{m}\hat{k}$$
Applying kinematics to find velocity at time $t$:
$$\vec{v}(t) = v_{0}\hat{i} + \left(\frac{eE_{0}t}{m}\right)\hat{k}$$
### Step 1: Calculating Velocity Magnitude and Final Wavelength
Find the magnitude of the updated velocity vector:
$$|\vec{v}| = \sqrt{v_{0}^{2} + \left(\frac{eE_{0}t}{m}\right)^{2}} = v_{0}\sqrt{1 + \frac{e^{2}E_{0}^{2}t^{2}}{m^{2}v_{0}^{2}}}$$
Substitute this into the wavelength equation:
$$\lambda' = \frac{h}{mv_{0}\sqrt{1 + \frac{e^{2}E_{0}^{2}t^{2}}{m^{2}v_{0}^{2}}}}$$
Since initial wavelength matches $\lambda_{0} = \frac{h}{mv_{0}}$, the expression simplifies to :
$$\lambda' = \frac{\lambda_{0}}{\sqrt{1 + \frac{e^{2}E_{0}^{2}t^{2}}{m^{2}v_{0}^{2}}}}$$
### Pattern Recognition
The perpendicular field increases the particle's overall velocity and momentum. Since wavelength is inversely proportional to momentum, it must decrease, which rules out options with a plus sign in the numerator.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Dual Nature of Radiation and Matter
Q8 (2025)
Which of the following phenomena can not be explained by wave theory of light? [cite: 52-53]
- Reflection of light
- Diffraction of light
- Refraction of light
- Compton effect
### Core Logic
* **Reflection, Refraction, and Diffraction** can all be fully explained using Huygens' principle and wave theory paths [cite: 54, 47, 66].
* **Compton effect** involves the scattering of an X-ray photon by an electron, demonstrating explicit momentum conversion. This process requires treating light strictly as localized particle packets (photons) and cannot be captured by continuous classical wave formulations.
### Step 1: Conclusion
Hence, the Compton effect is the correct answer as it relies entirely on the particle nature of electromagnetic waves.
### Pattern Recognition
Phenomena such as the Photoelectric Effect, Compton Scattering, and Blackbody Radiation serve as absolute foundational evidence for the particle/quantum layout of radiation, while Interference, Diffraction, and Polarization confirm wave configurations.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Dual Nature of Radiation and Matter
Q1286 (2025)
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:
- $\text{(A) is false but (R) is true.}$
- $\text{(A) is true but (R) is false.}$
- $\text{Both (A) and (R) are true and (R) is the correct explanation of (A).}$
- $\text{Both (A) and (R) are true but (R) is not the correct explanation of (A).}$
### Related Formula
$$eV_0 = h\nu - \phi_0$$
### Core Logic
Assertion (A) is true because applying a negative stopping potential decelerates the emitted photoelectrons and drops the output current down to zero.
Reason (R) is true because the stopping potential $V_0 = \left(\frac{h}{e}\right)\nu - \frac{\phi_0}{e}$ is linear with frequency $\nu$. However, the linearity of $V_0$ vs frequency does not explain the physical mechanism behind why a negative potential stops electron emission.
### Chapter Mix
Class 12 Physics: Dual Nature of Radiation and Matter
