Question

Electron beam when accelerated by a voltage of 10 kV, has a de-Broglie wavelength '$\lambda$'. If the voltage is increased to 20 kV then the de-Broglie wavelength associated with the electron beam would be

• A. $4\lambda$
• B. $2\lambda$
• C. $\frac{\lambda}{2}$
• D. $\frac{\lambda}{\sqrt{2}}$

Hint:

To halve the wavelength, the accelerating voltage must be increased four times.

Answer 1

The Correct Option is D

Step 1: de-Broglie Wavelength Formula
$\lambda = \frac{h}{\sqrt{2meV}}$. Since $h, m, e$ are constant, $\lambda \propto \frac{1}{\sqrt{V}}$.
Step 2: Set up Ratio
$\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{V_2}}$
Step 3: Calculation
$\frac{\lambda_2}{\lambda} = \sqrt{\frac{10}{20}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$
Step 4: Conclusion
The new wavelength is $\lambda/\sqrt{2}$.
Final Answer:(D)