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).
Q5 (2025)
An electron with mass $m$ with an initial velocity $(t = 0)$ $\vec{v} = v_0\hat{i}$ ($v_0 > 0$) enters a magnetic field $\vec{B} = B_0\hat{j}$ . If the initial de-Broglie wavelength at $t = 0$ is $\lambda_0$ then its value after time $t$ would be:
- $\frac{\lambda_0}{\sqrt{1 - \frac{e^2 B_0^2 t^2}{m^2}}}$
- $\frac{\lambda_0}{\sqrt{1 + \frac{e^2 B_0^2 t^2}{m^2}}}$
- $\lambda_0 \sqrt{1 + \frac{e^2 B_0^2 t^2}{m^2}}$
- $\lambda_0$
### Related Formula
1. Magnetic Force on a moving charge:
$$\vec{F} = q(\vec{v} \times \vec{B})$$
2. de-Broglie Wavelength:
$$\lambda = \frac{h}{p} = \frac{h}{m v}$$
where $p$ is the magnitude of momentum and $v$ is the speed.
### Core Logic
Since the magnetic force $\vec{F}$ is always perpendicular to the velocity $\vec{v}$ of the electron at any instant:
$$W = \int \vec{F} \cdot d\vec{r} = 0$$
By the work-energy theorem, since work done by the magnetic field is zero, the kinetic energy (and thus the speed $v$) of the electron remains constant throughout its motion.
Since speed $v = v_0$ (constant), the magnitude of momentum $p = m v$ remains constant over time.
Therefore, the de-Broglie wavelength remains unchanged:
$$\lambda(t) = \lambda_0$$
### Pattern Recognition
Sees: Charge entering purely magnetic field.
Trap: Resolving helical trajectories or cross products mathematically. Do not waste time computing components!
Shortcut: A magnetic field can ONLY change the direction of velocity, NEVER the magnitude (speed). Since de-Broglie wavelength depends solely on the magnitude of momentum ($p = mv$), it must remain constant.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Dual Nature of Radiation and Matter
Class 12 Physics: Moving Charges and Magnetism
Q20 (2025)
A monochromatic light is incident on a metallic plate having work function $\phi$. An electron, emitted normally to the plate from a point A with maximum kinetic energy, enters a constant magnetic field, perpendicular to the initial velocity of electron. The electron passes through a curve and hits back the plate at a point B. The distance between A and B is:
(Given: The magnitude of charge of an electron is $e$ and mass is $m$, $h$ is Planck's constant and $c$ is velocity of light. Take the magnetic field exists throughout the path of electron)
- $\sqrt{2m\left(\frac{hc}{\lambda} - \phi\right)} / eB$
- $\sqrt{m\left(\frac{hc}{\lambda} - \phi\right)} / eB$
- $\sqrt{8m\left(\frac{hc}{\lambda} - \phi\right)} / eB$
- $2 \sqrt{m\left(\frac{hc}{\lambda} - \phi\right)} / eB$
### Related Formula
$$K_{\max} = \frac{hc}{\lambda} - \phi$$
$$p = \sqrt{2m K_{\max}}$$
$$R = \frac{p}{eB}$$
$d = 2R$
### Core Logic
According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectron is:
$$K_{\max} = \frac{hc}{\lambda} - \phi$$
The momentum $p$ corresponding to this kinetic energy is:
$$p = \sqrt{2m K_{\max}} = \sqrt{2m \left(\frac{hc}{\lambda} - \phi\right)}$$
The electron is emitted normally to the plate and enters a perpendicular constant magnetic field $B$. It describes a circular arc (semicircle) and hits back the plate at point B. The distance between A and B is the diameter of this circular trajectory:
$$d_{AB} = 2R = 2 \left( \frac{p}{eB} \right) = \frac{2\sqrt{2m\left(\frac{hc}{\lambda} - \phi\right)}}{eB}$$
To align this with the options, move the factor of 2 inside the square root ($2 = \sqrt{4}$):
$$d_{AB} = \frac{\sqrt{4 \times 2m\left(\frac{hc}{\lambda} - \phi\right)}}{eB} = \frac{\sqrt{8m\left(\frac{hc}{\lambda} - \phi\right)}}{eB}$$$
### Step 1: Final Conclusion
The distance between points A and B is $\sqrt{8m\left(\frac{hc}{\lambda} - \phi\right)} / eB$.
### Pattern Recognition
When a particle is launched perpendicularly from a flat boundary into a perpendicular magnetic field, it describes a semicircle and exits/re-hits the boundary at a distance equal to the diameter $2R = 2\frac{p}{qB}$. Taking coefficients inside square roots converts $2 \sqrt{2x}$ to $\sqrt{8x}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Dual Nature of Matter and Radiation
Class 12 Physics: Moving Charges and Magnetism
Q21 (2025)
An electron is released from rest near an infinite non-conducting sheet of uniform charge density $-\sigma^{\prime}$. The rate of change of de-Broglie wavelength associated with the electron varies inversely as $n^{\text{th}}$ power of time. The numerical value of $n$ is
### Related Formula
$$\lambda = \frac{h}{p}$$
$$p = m v = m (at)$$
$$a = \frac{e E}{m} = \frac{e \sigma^{\prime}}{2m\varepsilon_0}$$
where,
$\lambda$ = de-Broglie wavelength
$p$ = linear momentum
$a$ = acceleration of the electron in the uniform electric field $E$
$t$ = time elapsed since release
### Core Logic
Since the electron starts from rest ($u = 0$), its velocity $v$ at any time $t$ is:
$v = at$
Thus, the momentum is $p = m v = m a t$.
Substitute this into the de-Broglie wavelength equation:
$$\lambda(t) = \frac{h}{m a t}$$
Now, compute the rate of change of wavelength with respect to time:
$$\frac{d\lambda}{dt} = \frac{d}{dt} \left( \frac{h}{ma} t^{-1} \right) = -\frac{h}{ma} t^{-2}$$
This shows that:
$$\left| \frac{d\lambda}{dt} \right| \propto \frac{1}{t^2}$$
Comparing this with the given statement (varies inversely as $n^{\text{th}}$ power of time):
$n = 2$
### Pattern Recognition
Sees: "Uniform electric field" + "de-Broglie wavelength rate of change" → Wavelength $\lambda \propto t^{-1}$.
Shortcut: Since $\lambda \propto \frac{1}{t}$, its derivative must scale as $\frac{d\lambda}{dt} \propto \frac{1}{t^2}$. Thus, $n = 2$ directly from basic power-rule differentiation! ✓
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Dual Nature of Matter and Radiation
Class 12 Physics: Electrostatics
Q4 (2025)
In an experiment with photoelectric effect, the stopping potential:
- $\text{increases with increase in the wavelength of the incident light}$
- $\text{increases with increase in the intensity of the incident light}$
- $\text{is } \left(\frac{1}{\mathrm{e}}\right) \text{ times the maximum kinetic energy of the emitted photoelectrons}$
- $\text{decreases with increase in the intensity of the incident light}$
### Related Formula
$$K_{\max} = h\nu - \phi = eV_s$$
where,
$K_{\max}$ = maximum kinetic energy of photoelectrons
$V_s$ = stopping potential
$e$ = fundamental electronic charge
### Core Logic
By definition, the stopping potential $V_s$ is the negative potential applied to stop the most energetic photoelectrons from reaching the collector electrode.
From Einstein's photoelectric equation:
$$eV_s = K_{\max} \implies V_s = \frac{K_{\max}}{e}$$
Thus, the stopping potential is exactly $\frac{1}{e}$ times the maximum kinetic energy of the emitted photoelectrons. It does not depend on the intensity of light.
### Pattern Recognition
Remember the primary features of the photoelectric effect:
- Stopping potential depends linearly on frequency, and inversely on wavelength.
- Intensity changes current, but has zero effect on stopping potential.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Dual Nature of Radiation and Matter
Q18 (2025)
A proton of mass $\mathrm{m_p}$ has same energy as that of a photon of wavelength $\lambda$ . If the proton is moving at non-relativistic speed, then ratio of its de Broglie wavelength to the wavelength of photon is.
- $\frac{1}{\mathrm{c}}\sqrt{\frac{2\mathrm{E}}{\mathrm{m}_{\mathrm{p}}}}$
- $\frac{1}{c} \sqrt{\frac{E}{m_p}}$
- $\frac{1}{c} \sqrt{\frac{E}{2 m_{p}}}$
- $\frac{1}{2c}\sqrt{\frac{E}{m_p}}$
### Core Logic
Let $\mathrm{E}$ represent the identical energy value shared by both particles:
$$\mathrm{E}_{\text{photon}} = \frac{\mathrm{hc}}{\lambda} = \mathrm{E}$$
$$\mathrm{E}_{\text{proton}} = \frac{\mathrm{p}^2}{2\mathrm{m}_{\mathrm{p}}} = \mathrm{E} \implies \mathrm{p} = \sqrt{2\mathrm{m}_{\mathrm{p}}\mathrm{E}}$$
Now, expressing the ratio of the proton's de Broglie wavelength to the photon's wavelength:
$$\frac{\lambda_{\text{proton}}}{\lambda_{\text{photon}}} = \frac{\mathrm{h}/\mathrm{p}}{\mathrm{hc}/\mathrm{E}} = \frac{\mathrm{h}/\sqrt{2\mathrm{m}_{\mathrm{p}}\mathrm{E}}}{\mathrm{hc}/\mathrm{E}}$$
$$\frac{\lambda_{\text{proton}}}{\lambda_{\text{photon}}} = \frac{\mathrm{E}}{\mathrm{c}\sqrt{2\mathrm{m}_{\mathrm{p}}\mathrm{E}}} = \frac{1}{\mathrm{c}} \sqrt{\frac{\mathrm{E}}{2\mathrm{m}_{\mathrm{p}}}}$$
### Step 1: Final Conclusion
The calculated ratio maps to option (3).
### Pattern Recognition
Combine the core formulas: $\lambda_{\text{matter}} = \frac{\mathrm{h}}{\sqrt{2\mathrm{mE}}}$ and $\lambda_{\text{light}} = \frac{\mathrm{hc}}{mathrm{E}}$. Dividing them smoothly yields the standard non-relativistic scaling ratio.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Dual Nature of Radiation and Matter
