Question

If the accelerating potential of a moving charged particle of \( \lambda \) de Broglie wavelength is increased by four times, then the de Broglie wavelength will be:

• A. \( \lambda/4 \)
• B. \( \lambda/2 \)
• C. \( 2\lambda \)
• D. \( 4\lambda \)

Hint:

Use \( \lambda = h/\sqrt{2mqV} \); the wavelength varies inversely with the square root of the accelerating potential.

Answer 1

The Correct Option is B

Step 1: Recall the de Broglie wavelength of a charged particle accelerated through a potential \( V \). The particle gains kinetic energy \( KE = qV \), so its momentum is \( p = \sqrt{2m\,qV} \).

Step 2: The de Broglie wavelength is \[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2m\,qV}}. \] This shows \( \lambda \propto \dfrac{1}{\sqrt{V}} \).

Step 3: Let the new potential be \( V' = 4V \). Then \[ \frac{\lambda'}{\lambda} = \sqrt{\frac{V}{V'}} = \sqrt{\frac{V}{4V}} = \frac{1}{2}. \]

Step 4: Therefore \( \lambda' = \dfrac{\lambda}{2} \), which is option 2. Options 3 and 4 wrongly assume \( \lambda \) grows with \( V \); option 1 uses the square of the factor instead of the square root.

\[\boxed{\lambda' = \frac{\lambda}{2}}\]