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 4.
Q1289 (2025)
If $\lambda$ and K are de Broglie Wavelength and kinetic energy, respectively, of a particle with constant mass. The correct graphical representation for the particle will be :-
### Related Formula
$\lambda = \frac{h}{\sqrt{2mK}}$
$$\lambda^2 = \left(\frac{h^2}{2m}\right) \frac{1}{K}$$
### Core Logic
Rearranging the de Broglie equation displays a parabolic relationship when evaluating squared attributes or corresponding axes coordinates. Given standard $\lambda$ vs $\frac{1}{\sqrt{K}}$ layout tracking, it exhibits an upward facing parabolic behavior matching option (2) layout.
### Chapter Mix
Class 12 Physics: Dual Nature of Radiation and Matter
Q41 (2024)
The de Broglie wavelengths of a proton and an $\alpha$ particle are $\lambda$ and $2\lambda$ respectively. The ratio of the velocities of proton and $\alpha$ particle will be:
- 1 : 8
- 1 : 2
- 4 : 1
- 8 : 1
### Related Formula
de Broglie wavelength relationship to velocity:
$$\lambda = \frac{h}{p} = \frac{h}{mv} \implies v = \frac{h}{m\lambda}$$
### Core Logic
Let mass of proton be $m_p$ and mass of $\alpha$ particle be $m_\alpha = 4m_p$.
Given wavelengths: $\lambda_p = \lambda$, $\lambda_\alpha = 2\lambda$.
Set up ratios:
$$\frac{v_p}{v_\alpha} = \frac{m_\alpha}{m_p} \times \frac{\lambda_\alpha}{\lambda_p}$$
### Step 1: Substitute Ratios
$$$\frac{v_p}{v_\alpha} = 4 \times \frac{2\lambda}{\lambda} = 4 \times 2 = 8$$$
Hence, the velocity ratio is $8:1$.
### Pattern Recognition
Remember the standard mass ratio: $m_\alpha \approx 4 m_p$. Inversely proportional components mean velocity amplifies significantly.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Dual Nature of Radiation and Matter
Q31 (2024)
Two sources of light emit with a power of $200\text{ W}$. The ratio of number of photons of visible light emitted by each source having wavelengths $300\text{ nm}$ and $500\text{ nm}$ respectively, will be:
- $1:5$
- $1:3$
- $5:3$
- $3:5$
### Related Formula
The power $P$ of a light source emitting $n$ photons per second of wavelength $\lambda$ is given by:
$$P = n \cdot \frac{hc}{\lambda}$$
where:
* $h$ is Planck's constant
* $c$ is the speed of light
### Core Logic
Since both light sources emit with the same power ($P = 200\text{ W}$), we can relate the number of photons emitted per second for each wavelength:
$$n_1 \frac{hc}{\lambda_1} = n_2 \frac{hc}{\lambda_2}$$
Cancelling out the constant terms $h$ and $c$, we get:
$$\frac{n_1}{\lambda_1} = \frac{n_2}{\lambda_2} \implies \frac{n_1}{n_2} = \frac{\lambda_1}{\lambda_2}$$
Thus, the ratio of the number of photons emitted is directly proportional to their wavelengths.
### Step 1: Substitute the Values
Given values:
* $\lambda_1 = 300\text{ nm}$
* $\lambda_2 = 500\text{ nm}$
Substituting these values into the ratio equation:
$$\frac{n_1}{n_2} = \frac{300}{500} = \frac{3}{5}$$
Thus, the ratio is $3:5$.
### Pattern Recognition
Shortcut: For equal power outputs, the photon emission rate $n$ is directly proportional to the wavelength $\lambda$. Therefore, ratio of photons $n_1 : n_2 = \lambda_1 : \lambda_2 = 300 : 500 = 3:5$ directly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Dual Nature of Radiation and Matter
Q44 (2024)
A convex lens of focal length $40\text{ cm}$ forms an image of an extended source of light on a photoelectric cell. A current $I$ is produced. The lens is replaced by another convex lens having the same diameter but focal length $20\text{ cm}$. The photoelectric current now is:
- $\frac{I}{2}$
- $4I$
- $2I$
- $I$
### Core Logic
Photoelectric current is directly proportional to the intensity of incident light, which depends on the amount of light energy intercepted. The amount of light energy collected by a lens depends strictly on its aperture diameter.
Since both lenses share the exact same diameter, they gather the same total light flux from the source and direct it onto the active photoelectric cell matrix. Thus, the total incident power is invariant.
### Step 1: Conclusion
Since the incident energy flux remains constant, the rate of emission of photoelectrons remains identical, meaning the current stays exactly $I$.
### Pattern Recognition
Focal length alters spatial image sizing metrics, but raw aperture dimensions rule total power intercept profiles in optical flux systems.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Dual Nature of Radiation and Matter, Ray Optics
Q43 (2024)
The de-Broglie wavelength of an electron is the same as that of a photon. If velocity of electron is $25\%$ of the velocity of light, then the ratio of K.E. of electron and K.E. of photon will be:
- $\frac{1}{1}$
- $\frac{1}{8}$
- $\frac{8}{1}$
- $\frac{1}{4}$
### Related Formula
For a photon, the energy (which is its kinetic energy) is:
$$E_p = \frac{h c}{\lambda_p} \implies \lambda_p = \frac{h c}{E_p}$$
For an electron, the de-Broglie wavelength related to its kinetic energy ($K_e$) is:
$$\lambda_e = \frac{h}{m_e v_e} = \frac{h v_e}{2 K_e}$$
since $K_e = \frac{1}{2} m_e v_e^2 \implies m_e v_e = \frac{2 K_e}{v_e}$.
### Core Logic
Given that the velocity of the electron is $25\%$ of the speed of light:
$$v_e = 0.25c = \frac{c}{4}$$
Substituting this into the electron's wavelength equation:
$$\lambda_e = \frac{h \left(\frac{c}{4}\right)}{2 K_e} = \frac{h c}{8 K_e}$$
### Step 1: Equate Wavelengths
We are given that the de-Broglie wavelengths are equal ($\lambda_p = \lambda_e$):
$$\frac{h c}{E_p} = \frac{h c}{8 K_e}$$
Cancelling $h c$ from both sides:
$$\frac{1}{E_p} = \frac{1}{8 K_e} \implies 8 K_e = E_p$$
### Step 2: Find the Energy Ratio
The ratio of the kinetic energy of the electron to that of the photon ($E_p$) is:
$$\frac{K_e}{E_p} = \frac{1}{8}$$
Therefore, the ratio is $1/8$.
### Pattern Recognition
A direct comparison between massive particle kinetic energy ($K = \frac{p^2}{2m} = \frac{p v}{2}$) and photon energy ($E = p c$) when they share the same momentum $p$ (same wavelength) yields a simple universally applicable rule: $\frac{K_{\text{particle}}}{E_{\text{photon}}} = \frac{v}{2c}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Dual Nature of Radiation and Matter
