Question

A proton accelerated through a potential difference \(V\) has a de-Broglie wavelength \(\lambda\). On doubling the potential difference, the de-Broglie wavelength of the proton

• A. Remains unchanged
• B. Becomes double
• C. Becomes \(4\) times
• D. Decreases by a factor of \(\dfrac{1}{\sqrt{2}}\)

Hint:

For a charged particle accelerated through potential \(V\), \[ \lambda=\frac{h}{\sqrt{2mqV}} \] Thus, \[ \lambda\propto \frac{1}{\sqrt{V}} \] If \(V\) becomes \(n\) times, the wavelength becomes \(\frac{1}{\sqrt{n}}\) times.

Answer 1

The Correct Option is D

Concept: The de-Broglie wavelength of a charged particle accelerated through a potential difference \(V\) is \[ \lambda=\frac{h}{\sqrt{2mqV}} \] Therefore, \[ \lambda\propto \frac{1}{\sqrt{V}} \]

Step 1:Write the proportionality relation. \[ \lambda\propto \frac{1}{\sqrt{V}} \] Initially, \[ \lambda_1=\lambda \] When the potential difference is doubled, \[ V_2=2V \]

Step 2: Find the new wavelength. \[ \frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{V_2}} \] \[ \frac{\lambda_2}{\lambda} = \sqrt{\frac{V}{2V}} \] \[ \frac{\lambda_2}{\lambda} = \frac{1}{\sqrt{2}} \] Hence, \[ \lambda_2 = \frac{\lambda}{\sqrt{2}} \]

Step 3: State the answer. \[ { \lambda_2=\frac{\lambda}{\sqrt{2}} } \] Therefore, the de-Broglie wavelength decreases by a factor of \[ {\frac{1}{\sqrt{2}}} \] Hence, the correct option is \[ {(D)} \]