Question

An electron accelerated by a potential difference '$V$' has de-Broglie wavelength '$\lambda$'. If the electron is accelerated by a potential difference '$9 \text{ V}$', its de-Broglie wavelength will be}

• A. $\frac{\lambda}{4.5}$
• B. $\frac{\lambda}{3}$
• C. $\frac{\lambda}{2}$
• D. $\lambda$

Hint:

Potential increases by 9 times $\rightarrow$ Wavelength decreases by $\sqrt{9} = 3$ times.

Answer 1

The Correct Option is B

Step 1: Concept
The de-Broglie wavelength of an accelerated electron is inversely proportional to the square root of the accelerating potential ($\lambda \propto 1/\sqrt{V}$).

Step 2: Meaning
$\lambda_1 = k/\sqrt{V}$ and $\lambda_2 = k/\sqrt{9V}$.

Step 3: Analysis
$\lambda_2 / \lambda_1 = \sqrt{V} / \sqrt{9V} = 1/3$.

Step 4: Conclusion
$\lambda_2 = \lambda/3$. Final Answer: (B)