Question

An electron of mass '\(m\)' accelerated through a potential difference '\(V\)' has the de Broglie wavelength associated with it as '\(\lambda\)'. If a proton of a mass '\(M\)' is accelerated through the same potential difference '\(V\)', the de Broglie wavelength associated with it will be

• A. \(\left(\sqrt{\frac{m}{M}}\right)\lambda\)
• B. \(\left(\sqrt{\frac{M}{m}}\right)\lambda\)
• C. \(\frac{\lambda M}{m}\)
• D. \(\frac{M}{\lambda m}\)

Hint:

For a charge accelerated through potential V, the de Broglie wavelength depends inversely on the square root of its mass.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
A charge q accelerated through V gains \(qV = \dfrac{p^{2}}{2m}\), so \(p = \sqrt{2mqV}\).

Step 2: de Broglie wavelength
\(\lambda = \dfrac{h}{p} = \dfrac{h}{\sqrt{2mqV}}\).

Step 3: Compare electron and proton
Both have the same charge magnitude and the same V, so \(\lambda \propto \dfrac{1}{\sqrt{m}}\).
Thus \(\dfrac{\lambda_{p}}{\lambda} = \sqrt{\dfrac{m}{M}}\).

Final Answer:
The proton's wavelength is the square root of m over M times lambda. \[ \boxed{\text{(A) }\left(\sqrt{\dfrac{m}{M}}\right)\lambda} \]