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: Understanding the Question:
The de-Broglie wavelength of a charged particle depends on its kinetic energy, which is determined by the accelerating potential difference $V$.
Step 2: Key Formula or Approach:
\[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2meV}} \]
Thus, $\lambda \propto \frac{1}{\sqrt{V}}$.
Step 3: Detailed Explanation:
Let $V_1 = 10$ kV and $V_2 = 20$ kV.
Wavelength ratio:
\[ \frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{V_2}} \]
\[ \frac{\lambda_2}{\lambda} = \sqrt{\frac{10}{20}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \]
\[ \lambda_2 = \frac{\lambda}{\sqrt{2}} \]
Step 4: Final Answer:
The new wavelength is $\frac{\lambda}{\sqrt{2}}$.