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 6.
Q48 (2026)
A particle having electric charge $3 \times 10^{-19} \text{ C}$ and mass $6 \times 10^{-27} \text{ kg}$ is accelerated by applying an electric potential of $1.21 \text{ V}$. Wavelength of the matter wave associated with the particle is $\alpha \times 10^{-12} \text{ m}$. The value of $\alpha$ is ________.
(Take Planck's constant $= 6.6 \times 10^{-34} \text{ J}\cdot\text{s}$)
### Related Formula
$$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$$
$K = qV$
### Core Logic
For a charged particle accelerated through a potential difference $V$, the de Broglie wavelength is:
$$\lambda = \frac{h}{\sqrt{2mqV}}$$
### Step 1: Value Substitution
$$\lambda = \frac{6.6 \times 10^{-34}}{\sqrt{2 \times 6 \times 10^{-27} \times 3 \times 10^{-19} \times 1.21}}$$
$$\lambda = \frac{6.6 \times 10^{-34}}{\sqrt{36 \times 10^{-46} \times 1.21}}$$
### Step 2: Arithmetic Evaluation
Evaluate the square root:
$$\sqrt{36} \times \sqrt{1.21} \times \sqrt{10^{-46}} = 6 \times 1.1 \times 10^{-23} = 6.6 \times 10^{-23}$$
$$\lambda = \frac{6.6 \times 10^{-34}}{6.6 \times 10^{-23}}$$
$$\lambda = 10^{-11} \text{ m}$$
### Step 3: Final Conclusion
We need to express this in the form $\alpha \times 10^{-12} \text{ m}$:
$$10^{-11} \text{ m} = 10 \times 10^{-12} \text{ m}$$
Therefore, $\alpha = 10$.
### Pattern Recognition
The expression inside the radical always evaluates cleanly in JEE. $2 \times 6 \times 3 = 36$ and $1.21 = (1.1)^2$ are engineered to perfectly cancel the $6.6$ in the numerator.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Dual Nature of Radiation and Matter
Q44 (2026)
Light is incident on a metallic plate having work function $110 \times 10^{-20}$ J. If the produced photoelectrons have zero kinetic energy then the angular frequency of the incident light is ____ rad/s. ($h = 6.63 \times 10^{-34}$ J.s)
- $1.04 \times 10^{16}$
- $1.04 \times 10^{13}$
- $1.66 \times 10^{16}$
- $1.66 \times 10^{15}$
### Related Formula
$\phi = h v$
$$\omega = 2\pi v = \frac{2\pi \phi}{h}$$
### Core Logic
Since kinetic energy of photoelectrons is zero ($K_{\text{max}} = 0$), incident photon energy equals work function $\phi$:
$$h v = \phi \implies v = \frac{\phi}{h}$$
Calculating angular frequency $\omega$:
$$\omega = 2\pi v = \frac{2\pi \phi}{h} = \frac{2 \times 3.14 \times 110 \times 10^{-20}}{6.63 \times 10^{-34}}$$
$$\omega = \frac{690.8 \times 10^{-20}}{6.63 \times 10^{-34}} \approx 1.04 \times 10^{16} \mathrm{~rad/s}$$
### Step 1: Final Conclusion
The angular frequency of incident light is $1.04 \times 10^{16} \mathrm{~rad/s}$.
### Pattern Recognition
Zero kinetic energy $\implies h\nu = \phi$. Convert linear frequency $\nu$ to angular frequency $\omega = 2\pi \phi / h$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Dual Nature of Radiation and Matter
Q30 (2026)
The de Broglie wavelength of an oxygen molecule at $27^{\circ}C$ is $x \times 10^{-12}$ m. The value of x is (take Planck's constant = $6.63 \times 10^{-34}\text{ J.s}$, Boltzmann constant = $1.38 \times 10^{-23}\text{ J/K}$, mass of oxygen. Molecule = $5.31 \times 10^{-26}\text{ kg}$).
- $26$
- $24$
- $30$
- $20$
### Related Formula
$$\lambda = \frac{h}{\sqrt{2mK}}$$
$$K = \frac{3}{2}kT$$
$$\lambda = \frac{h}{\sqrt{3mkT}}$$
### Step 1: Substitute Values
Given values:
$h = 6.63 \times 10^{-34} \text{ J}\cdot\text{s}$
$m = 5.31 \times 10^{-26} \text{ kg}$
$k = 1.38 \times 10^{-23} \text{ J/K}$
$T = 27^{\circ}\text{C} = 27 + 273 = 300 \text{ K}$
Substitute these into the wavelength equation:
$$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{3 \times 5.31 \times 10^{-26} \times 1.38 \times 10^{-23} \times 300}}$$
### Step 2: Simplify Calculation
$$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{3 \times 300 \times 5.31 \times 1.38 \times 10^{-49}}}$$
$$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{900 \times 7.3278 \times 10^{-49}}}$$
$$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{6595.02 \times 10^{-49}}}$$
$$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{659.502 \times 10^{-48}}}$$
$$\lambda = \frac{6.63 \times 10^{-34}}{25.68 \times 10^{-24}} \approx 0.258 \times 10^{-10} \text{ m}$$
### Step 3: Match with Format
$$\lambda = 25.8 \times 10^{-12} \text{ m}$$
Rounding to nearest integer, $x = 26$.
### Pattern Recognition
Sees: "de Broglie wavelength" + "gas molecule" + "temperature" → Immediately use $\lambda = h/\sqrt{3mkT}$. Watch for temperature conversion to Kelvin.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Dual Nature of Radiation and Matter
Class 11 Physics: Kinetic Theory
Q33 (2026)
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.
$(\mathrm{h} = 6.63\times 10^{-34}\mathrm{J.s},\mathrm{e} = 1.6\times 10^{-19}\mathrm{C},\mathrm{c} = 3\times 10^{8}\mathrm{m / s})$
- $2.9 \times 10^{-8}$
- $2.2 \times 10^{-8}$
- $3.1 \times 10^{-7}$
- $2.5 \times 10^{-7}$
### Related Formula
$$eV_s = \frac{hc}{\lambda} - \phi$$
### Core Logic
For the first light:
$$e(3.2) = \frac{hc}{\lambda} - \phi \quad \dots (1)$$
For the second light (wavelength $2\lambda$):
$$e(0.7) = \frac{hc}{2\lambda} - \phi \quad \dots (2)$$
{{SOL_IMG1}}
### Step 1: Solving the Equation
Subtract Equation (2) from Equation (1):
$$e(3.2) - e(0.7) = \left( \frac{hc}{\lambda} - \phi \right) - \left( \frac{hc}{2\lambda} - \phi \right)$$
$$e(2.5) = \frac{hc}{2\lambda}$$
### Step 2: Calculate Wavelength
$$2.5 = \left(\frac{hc}{e}\right) \left(\frac{1}{2\lambda}\right)$$
Using $\frac{hc}{e} \approx 12400 \text{ eV}\cdot\text{\AA}$:
$$2.5 = \frac{12400}{2\lambda}$$
$$\lambda = \frac{12400}{5} \text{ \AA}$$
$$\lambda = 2480 \text{ \AA} = 2.48 \times 10^{-7} \text{ m}$$
### Pattern Recognition
When you have two states of photoelectric effect for the same metal, subtracting the two stopping potential equations instantly eliminates the work function $\phi$. Convert $hc/e$ directly to $12400 \text{ eV}\cdot\text{\AA}$ to expedite calculation.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Dual Nature of Radiation and Matter
Q47 (2026)
The ratio of de Broglie wavelength of a deutron with kinetic energy E to that of an alpha particle with kinetic energy 2E, is $n : 1$. The value of n is ____.
(Assume mass of proton = mass of neutron)
### Related Formula
$$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m \cdot KE}}$$
### Core Logic
Set up the ratio of wavelengths for the deuteron and the alpha particle based on their given kinetic energies and masses. Let mass of proton/neutron be $m$. Deuteron mass is $2m$, Alpha particle mass is $4m$.
### Step 1: Ratio setup
$$\frac{\lambda_d}{\lambda_\alpha} = \sqrt{\frac{m_\alpha \cdot KE_\alpha}{m_d \cdot KE_d}}$$
### Step 2: Value Substitution
Substitute $m_\alpha = 4m$, $m_d = 2m$, $KE_\alpha = 2E$, and $KE_d = E$:
$$= \sqrt{\frac{4m \cdot 2E}{2m \cdot E}}$$
$$= \sqrt{\frac{8mE}{2mE}} = \sqrt{4} = 2$$
So, the ratio is $2 : 1$. Therefore, $n = 2$.
### Pattern Recognition
Memorize mass ratios for common particles: proton ($m$), deuteron ($2m$), alpha ($4m$). Substitute directly into inverse-sqrt formula for $\lambda$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Dual Nature of Radiation and Matter
