Q105 / 150 · MHT CET 2026 (13 April Shift 2) · Physics
An electron accelerated through a potential difference $V_1$ has a de-Broglie wavelength of $\lambda$. When the potential is changed to $V_2$, its de-Broglie wavelength increases to $2\lambda$. The value of $(V_1/V_2)$ is equal to
- A$3:1$
- B$9:4$
- C$3:2$
- Correct answerD$4:1$
Solution
The de-Broglie wavelength of an electron accelerated through a potential difference $V$ is given by $\lambda = \dfrac{h}{\sqrt{2meV}}$.
This implies $\lambda \propto \dfrac{1}{\sqrt{V}}$, or $V \propto \dfrac{1}{\lambda^2}$.
For the first case, $V_1 \propto \dfrac{1}{\lambda^2}$.
For the second case, $V_2 \propto \dfrac{1}{(2\lambda)^2} = \dfrac{1}{4\lambda^2}$.
Taking the ratio of the two potentials:
$\dfrac{V_1}{V_2} = \dfrac{\dfrac{1}{\lambda^2}}{\dfrac{1}{4\lambda^2}} = 4$
Thus, $\dfrac{V_1}{V_2} = 4:1$.
Answer: $4:1$
This implies $\lambda \propto \dfrac{1}{\sqrt{V}}$, or $V \propto \dfrac{1}{\lambda^2}$.
For the first case, $V_1 \propto \dfrac{1}{\lambda^2}$.
For the second case, $V_2 \propto \dfrac{1}{(2\lambda)^2} = \dfrac{1}{4\lambda^2}$.
Taking the ratio of the two potentials:
$\dfrac{V_1}{V_2} = \dfrac{\dfrac{1}{\lambda^2}}{\dfrac{1}{4\lambda^2}} = 4$
Thus, $\dfrac{V_1}{V_2} = 4:1$.
Answer: $4:1$
Hence, the correct answer is option: 'D'.