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 5.
Q33 (2024)
For the photoelectric effect, the maximum kinetic energy $\left(\mathrm{E}_{\mathrm{k}}\right)$ of the photoelectrons is plotted against the frequency ($\nu$) of the incident photons as shown in figure. The slope of the graph gives
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- $\text{Ratio of Planck's constant to electric charge}$
- $\text{Work function of the metal}$
- $\text{Charge of electron}$
- $\text{Planck's constant}$
### Related Formula
$$\mathrm{K.E.}_{\max} = h\nu - \phi$$
### Core Logic
From Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is given by $E_k = h\nu - \phi$, where $h$ is Planck's constant, $\nu$ is the incident frequency, and $\phi$ is the work function.
This represents a straight line equation of the form $y = mx + c$, where $y = E_k$ and $x = \nu$.
### Step 1: Identify the Slope
Comparing $E_k = h\nu - \phi$ with $y = mx + c$, we get the slope:
$$m = \tan \theta = h$$
Thus, the slope of the graph gives Planck's constant.
### Pattern Recognition
If the y-axis is $K.E.$, the slope is $h$. If the y-axis is Stopping Potential $V_0$, the equation is $eV_0 = h\nu - \phi \implies V_0 = (h/e)\nu - (\phi/e)$, and the slope would be $h/e$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Dual Nature of Radiation and Matter
Q35 (2024)
The work function of a substance is $3.0 \mathrm{~eV}$. The longest wavelength of light that can cause the emission of photoelectrons from this substance is approximately:
- $215 \mathrm{~nm}$
- $414 \mathrm{~nm}$
- $400 \mathrm{~nm}$
- $200 \mathrm{~nm}$
### Related Formula
$$E = \frac{hc}{\lambda}$$
$$\lambda_{th} = \frac{1240 \text{ eV}\cdot\text{nm}}{\Phi}$$
### Core Logic
For photoelectric emission to occur, the incident energy must be greater than or equal to the work function ($W_e$ or $\Phi$).
The longest wavelength corresponds to the minimum energy required, which is exactly the work function.
$$\lambda \le \frac{hc}{W_e}$$
### Step 1: Calculate Wavelength
Substitute the given values using the convenient constant $hc \approx 1240 \mathrm{~eV \cdot nm}$:
$$\lambda \le \frac{1240 \mathrm{~eV \cdot nm}}{3.0 \mathrm{~eV}}$$
$$\lambda \le 413.33 \mathrm{~nm}$$
The maximum wavelength (longest wavelength) is:
$$\lambda_{\max} \approx 414 \mathrm{~nm}$$
### Pattern Recognition
The standard shortcut $E(\text{eV}) = 1240 / \lambda(\text{nm})$ rapidly converts work function to threshold wavelength without dealing with standard SI unit conversions.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Dual Nature of Radiation and Matter
Q41 (2024)
In a photoelectric effect experiment a light of frequency 1.5 times the threshold frequency is made to fall on the surface of photosensitive material. Now if the frequency is halved and intensity is doubled, the number of photo electrons emitted will be:
- Doubled
- Quadrulated
- Zero
- Halved
### Related Formula
For photoelectric emission to occur:
$f \ge f_0$
where $f$ is incident frequency and $f_0$ is the threshold frequency.
### Core Logic
Initially, the frequency is $f_1 = 1.5 f_0$. Since $f_1 > f_0$, emission happens.
Then, the frequency is halved: $f_2 = \frac{1.5 f_0}{2} = 0.75 f_0$.
### Step 1: Check Threshold Condition
Since $f_2 = 0.75 f_0$, we see that $f_2 < f_0$.
The incident light no longer has enough energy per photon to overcome the work function, regardless of how intense the light is.
### Step 2: Conclusion
Because the threshold condition fails, emission completely stops. The number of photoelectrons emitted is zero.
### Pattern Recognition
Always check the frequency threshold first in photoelectric questions. Intensity adjustments are irrelevant traps if $f < f_0$. No emission occurs below threshold frequency.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Dual Nature of Radiation and Matter
Q50 (2024)
When a metal surface is illuminated by light of wavelength $\lambda$, the stopping potential is $8\mathrm{\ V}$. When the same surface is illuminated by light of wavelength $3\lambda$, stopping potential is $2\mathrm{\ V}$. The threshold wavelength for this surface is:
- $5\lambda$
- $3\lambda$
- $9\lambda$
- $4.5\lambda$
### Related Formula
$$\frac{hc}{\lambda} = \phi + K_{\text{max}}$$
$$K_{\text{max}} = eV_0$$
$$\phi = \frac{hc}{\lambda_0}$$
### Core Logic
Using Einstein's Photoelectric equation for the two cases:
Case 1 (Wavelength $\lambda$, Stopping Potential $8\mathrm{\,V}$):
$$\frac{hc}{\lambda} = \frac{hc}{\lambda_0} + 8e \quad \dots \dots (\text{i})$$
Case 2 (Wavelength $3\lambda$, Stopping Potential $2\mathrm{\,V}$):
$$\frac{hc}{3\lambda} = \frac{hc}{\lambda_0} + 2e \quad \dots \dots (\text{ii})$$
### Step 2: Solving the Equations
Multiply equation (ii) by 4 to eliminate $e$:
$$\frac{4hc}{3\lambda} = \frac{4hc}{\lambda_0} + 8e$$
Equating this to equation (i):
$$\frac{hc}{\lambda} - \frac{hc}{\lambda_0} = \frac{4hc}{3\lambda} - \frac{4hc}{\lambda_0}$$
Divide entirely by $hc$:
$$\frac{1}{\lambda} - \frac{1}{\lambda_0} = \frac{4}{3\lambda} - \frac{4}{\lambda_0}$$
$$\frac{4}{\lambda_0} - \frac{1}{\lambda_0} = \frac{4}{3\lambda} - \frac{1}{\lambda}$$
$$\frac{3}{\lambda_0} = \frac{4 - 3}{3\lambda}$$
$$\frac{3}{\lambda_0} = \frac{1}{3\lambda}$$
$$\lambda_0 = 9\lambda$$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Dual Nature Of Radiation And Matter
Q37 (2026)
A light wave described by $E = 60[\sin(3 \times 10^{15}t) + \sin(12 \times 10^{15}t)]$ (in SI units) falls on a metal surface of work function 2.8 eV. The maximum kinetic energy of ejected photoelectron is (approximately) ____ eV. ($h = 6.6 \times 10^{-34}\text{ J}\cdot\text{s}$. and $e = 1.6 \times 10^{-19}\text{C}$)
- 5.1
- 3.8
- 6.0
- 7.8
### Related Formula
$$K_{\text{max}} = h\nu_{\text{max}} - \phi_0$$
$$v = \frac{\omega}{2\pi}$$
### Core Logic
The light wave consists of two frequencies governed by $\omega_1$ and $\omega_2$.
$\omega_1 = 3 \times 10^{15}\text{ rad/s}$
$\omega_2 = 12 \times 10^{15}\text{ rad/s}$
The maximum kinetic energy of ejected photoelectrons will be determined by the highest frequency photon, which corresponds to $\omega_2 = 12 \times 10^{15}\text{ rad/s}$.
### Step 1: Calculate Photon Energy
Frequency $\nu_{\text{max}} = \frac{\omega_2}{2\pi} = \frac{12 \times 10^{15}}{2 \times 3.14} \approx 1.91 \times 10^{15}\text{ Hz}$.
Energy of this photon:
$$E_{\text{photon}} = h\nu = (6.6 \times 10^{-34}) \times (1.91 \times 10^{15}) = 1.26 \times 10^{-18}\text{ J}$$
Convert this energy to eV:
$$E_{\text{max}} = \frac{1.26 \times 10^{-18}}{1.6 \times 10^{-19}} \approx 7.87\text{ eV} \approx 7.9\text{ eV}$$
### Step 2: Calculate Maximum Kinetic Energy
Using Einstein's photoelectric equation:
$$K_{\text{max}} = E_{\text{max}} - \phi_0$$
$$K_{\text{max}} = 7.9 - 2.8 = 5.1\text{ eV}$$
### Pattern Recognition
When a wave has multiple frequency components ($E = E_1\sin\omega_1 t + E_2\sin\omega_2 t$), the $K_{\text{max}}$ is always strictly determined by the highest frequency (highest energy) component. Ignore the lower frequency terms.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Dual Nature of Radiation and Matter
