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 2.
Q15 (2025)
The work function of a metal is $3\mathrm{~eV}$. The color of the visible light that is required to cause emission of photoelectrons is:
- Green
- Blue
- Red
- Yellow
### Related Formula
Einstein's Photoelectric Equation:
$$K_{\text{max}} = h\nu - \phi = \frac{hc}{\lambda} - \phi$$
For emission to occur, the photon energy must exceed the work function:
$$E_{\text{photon}} > \phi \implies \lambda < \lambda_{\text{threshold}} = \frac{hc}{\phi}$$
Useful shortcut: $hc \approx 1240\mathrm{~eV\cdot nm}$.
### Core Logic
Let's find the threshold wavelength $\lambda_{\text{threshold}}$ for a work function of $\phi = 3\mathrm{~eV}$:
$$\lambda_{\text{threshold}} = \frac{1240\mathrm{~eV\cdot nm}}{3\mathrm{~eV}} \approx 413.3\mathrm{~nm}$$
To cause photoelectric emission, the wavelength of the incident light must be shorter than this threshold:
$$\lambda < 413.3\mathrm{~nm}$$
### Step 1: Comparing Visible Spectrum Colors
Let's look at the standard approximate wavelength ranges for visible light:
- **Red**: $620 - 750\mathrm{~nm}$ (Energy $\approx 1.65 - 2.0\mathrm{~eV}$)
- **Yellow**: $570 - 590\mathrm{~nm}$ (Energy $\approx 2.1 - 2.2\mathrm{~eV}$)
- **Green**: $495 - 570\mathrm{~nm}$ (Energy $\approx 2.2 - 2.5\mathrm{~eV}$)
- **Blue**: $450 - 495\mathrm{~nm}$ (At lower end, approaching violet down to $380\mathrm{~nm}$; Energy $\approx 2.5 - 3.3\mathrm{~eV}$)
Only **Blue** light contains wavelengths extending below $413.3\mathrm{~nm}$ (high enough photon energy to surpass $3\mathrm{~eV}$). Therefore, blue light is required to cause photoelectric emission from this metal.
### Pattern Recognition
Remember standard photon energies of visible colors: Blue/Violet photons have higher energy (typically $> 2.8\mathrm{~eV}$), while Red/Yellow photons have much lower energy ($< 2.2\mathrm{~eV}$). For a higher work function like $3\mathrm{~eV}$, only highly energetic blue/violet light can succeed.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Dual Nature of Radiation and Matter
Q902 (2025)
A small mirror of mass $m$ is suspended by a massless thread of length $l$. Then the small angle through which the thread will be deflected when a short pulse of laser of energy $E$ falls normal on the mirror ($c = $ speed of light in vacuum and $g = $ acceleration due to gravity)
- $\theta = \frac{3E}{4mcsqrt{gl}}$
- $\theta = \frac{E}{mcsqrt{gl}}$
- $\theta = \frac{E}{2mcsqrt{gl}}$
- $\theta = \frac{2E}{mcsqrt{gl}}$
### Related Formula
Force due to a completely reflecting beam:
$$F = \frac{2P}{c} = \frac{2}{c}\frac{dE}{dt}$$
Change in momentum:
$$\Delta p = mv = \int F \, dt = \frac{2E}{c}$$
Work-Energy Theorem or Conservation of Mechanical Energy for small deflections:
$$gl\left(2\sin^2\frac{\theta}{2}\right) = \frac{v^2}{2}$$
For small angles $\sin\frac{\theta}{2} \approx \frac{\theta}{2}$.
### Core Logic
Assuming perfect normal reflection from the mirror surface, the pulse imparts a momentum impulse of $\frac{2E}{c}$ to the mass. This provides an initial velocity $v$ to the mirror. The mirror then swings up to a maximum angle $\theta$ where kinetic energy converts entirely to gravitational potential energy.
{{SOL_IMG1}}
### Step 1: Calculate Initial Velocity
From momentum change:
$$m(v - 0) = \frac{2E}{c} \implies v = \frac{2E}{mc}$$
{{SOL_IMG2}}
### Step 2: Relate to Angular Deflection
Using conservation of mechanical energy:
$$mgl(1 - \cos\theta) = \frac{1}{2}mv^2$$
$$gl\left(2\sin^2\frac{\theta}{2}\right) = \frac{v^2}{2}$$
Since $\theta$ is very small, $\sin\frac{\theta}{2} \approx \frac{\theta}{2}$:
$$gl\left(2\left(\frac{\theta}{2}\right)^2\right) = \frac{v^2}{2} \implies gl\frac{\theta^2}{2} = \frac{v^2}{2} \implies gl\theta^2 = v^2$$
### Step 3: Solve for Theta
Substitute $v = \frac{2E}{mc}$ into the expression:
$$gl\theta^2 = \left(\frac{2E}{mc}\right)^2 = \frac{4E^2}{m^2c^2}$$
$$\theta^2 = \frac{4E^2}{m^2c^2gl} \implies \theta = \frac{2E}{mc\sqrt{gl}}$$
### Pattern Recognition
This is a standard ballistic pendulum problem where the impulse is delivered by radiation pressure. Perfect reflection means momentum transfer is double the incident momentum $\left(2\cdot \frac{E}{c}\right)$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Dual Nature of Radiation and Matter
Class 11 Physics: System of Particles and Rotational Motion
Q10 (2025)
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason 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
- Both A and R are true but R is NOT the correct explanation of A
- A is false but R is true
- A is true but R is false
- Both A and R are true and R is the correct explanation of A
### Related Formula
Einstein's photoelectric equation:
$$
eV_s = h
u - \phi$$
where:
* $V_s$ = stopping potential
* $
u$ = frequency of light
* $\phi$ = work function
* Intensity formula: $I = \frac{n h
u}{A \cdot t}$ (where $n$ is rate of photons).
### Core Logic
* **Assertion Analysis:** Stopping potential $V_s$ depends strictly linearly on frequency $
u$ and work function $\phi$. It is completely independent of the beam intensity. Therefore, Assertion A is **false**.
* **Reason Analysis:** Intensity tracks the flux counts of photons per second. Increasing intensity drives up the quantum count of ejected charges, given $
u >
u_0$. Thus, Reason R is **true**.
### Pattern Recognition
Stopping Potential $\leftrightarrow$ Frequency/Energy characteristic.
Photo-current / Emission Rate $\leftrightarrow$ Photon Intensity/Flux counts.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Dual Nature of Radiation and Matter
Q5 (2025)
A photo-emissive substance is illuminated with a radiation of wavelength $\lambda_{i}$ so that it releases electrons with de-Broglie wavelength $\lambda_{c}$ The longest wavelength of radiation that can emit photoelectron is $\lambda_{0}$.
Expression for de-Broglie wavelength is given by:
(m: mass of the electron, h: Planck's constant and c: speed of light) [cite: 39, 41]
- $\lambda_{e}=\sqrt{\frac{h}{2mc(\frac{1}{\lambda_{i}}-\frac{1}{\lambda_{o}})}}$ [cite: 46]
- $\lambda_{c}=\sqrt{\frac{h\lambda_{0}}{2mc}}$ [cite: 47]
- $\lambda_{e}=\frac{h}{\sqrt{2mc(\frac{1}{\lambda_{i}}-\frac{1}{\lambda_{o}})}}$ [cite: 48]
- $\lambda_{c}=\sqrt{\frac{h\lambda_{i}}{2mc}}$ [cite: 49]
### Related Formula
$$\text{K.E} = E - W$$ [cite: 681]
$$\lambda_{e} = \frac{h}{\sqrt{2m\text{K.E}}}, \quad E = \frac{hc}{\lambda_{i}}, \quad W = \frac{hc}{\lambda_{0}}$$ [cite: 682]
### Core Logic
From the de-Broglie wavelength relationship, squaring both sides yields: [cite: 682]
$$\frac{h^2}{2m\lambda_{e}^2} = \text{K.E} = \frac{hc}{\lambda_{i}} - \frac{hc}{\lambda_{0}} = hc\left(\frac{1}{\lambda_{i}} - \frac{1}{\lambda_{0}}\right)$$ [cite: 682]
Simplifying for $\lambda_{e}$: [cite: 682]
$$\lambda_{e}^2 = \frac{h^2}{2mhc\left(\frac{1}{\lambda_{i}} - \frac{1}{\lambda_{0}}\right)} = \frac{h}{2mc\left(\frac{1}{\lambda_{i}} - \frac{1}{\lambda_{0}}\right)}$$
$$\lambda_{e} = \sqrt{\frac{h}{2mc\left(\frac{1}{\lambda_{i}} - \frac{1}{\lambda_{0}}\right)}}$$ [cite: 682]
### Pattern Recognition
Einstein's photoelectric equation relates kinetic energy linearly to $\frac{1}{\lambda}$ parameters[cite: 681, 682]. Combining this directly into the momentum term $\sqrt{2mK}$ under the Planck constant yields the standard reciprocal root difference layout[cite: 682].
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Dual Nature of Radiation and Matter
Q12 (2025)
In photoelectric effect, the stopping potential ($V_{0}$) v/s frequency ($
u$) curve is plotted. ($h$ is the Planck's constant and $\phi_0$ is work function of metal)
(A) $V_{0}$ v/s $
u$ is linear
(B) The slope of $V_{0}$ v/s $
u$ curve = $\frac{\phi_{0}}{h}$
(C) $h$ constant is related to the slope of $V_{0}$ v/s $
u$ line
(D) The value of electric charge of electron is not required to determine $h$ using the $V_{0}$ v/s $
u$ curve.
(E) The work function can be estimated without knowing the value of $h$.
Choose the correct answer from the options given below :
- (A), (B) and (C) only
- (C) and (D) only
- (A), (C) and (E) only
- (D) and (E) only
### Related Formula
$$h
u = \phi_0 + K_{\text{\max}} = \phi_0 + eV_0$$
$$V_0 = \left(\frac{h}{e}
ight)
u - \frac{\phi_0}{e}$$
### Core Logic
Analyzing each statement:
- (A) $V_0 \text{ v/s }
u$ is a straight line equation ($y = mx + c$). True.
- (B) Slope is $\frac{h}{e}$, not $\frac{\phi_0}{h}$. False.
- (C) The slope involves Planck's constant $h$. True.
- (D) To find $h$ from the slope ($m = h/e$), you must multiply the slope by the electronic charge $e$. Thus, $e$ is required. False.
- (E) The $
u$-intercept occurs when $V_0 = 0 \implies
u_{\text{th}} = \frac{\phi_0}{h}$. The $y$-intercept is $-\phi_0/e$. Therefore, we can find $\phi_0$ from the intercepts knowing only $e$ or by looking at the scale parameters carefully, without explicitly knowing $h$. True.
Hence, statements (A), (C), and (E) are correct.
### Pattern Recognition
Einstein's photoelectric equation yields a straight-line plot for $V_0$ vs $
u$ where slope is universally $\frac{h}{e}$ for all metals.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Dual Nature of Radiation and Matter
