A proton accelerated through a potential difference V has a de-Broglie wavelength λ. On doubling the potential difference, the de-Broglie wavelength of the proton-
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → decreases by a factor of 1/√2
The de Broglie wavelength is \( \lambda = h/p \). A proton accelerated through a potential difference V gains kinetic energy eV, so its momentum is \( p = \sqrt{2meV} \).
\( \displaystyle \lambda = \frac{h}{\sqrt{2meV}} \Rightarrow \lambda \propto \frac{1}{\sqrt{V}} \)
\( \displaystyle \frac{\lambda'}{\lambda} = \sqrt{\frac{V}{2V}} = \frac{1}{\sqrt{2}} \)
\( \displaystyle \lambda' = \frac{\lambda}{\sqrt{2}} \)
Doubling the potential difference makes the wavelength 1/√2 times its earlier value, i.e. it decreases; this is the change Option (4) describes.
Note: It seems that there is a mistake in the original NTA paper, which prints “decreases by a factor of 1/√2” when it should read “decreases by a factor of √2” (the wavelength becomes 1/√2 times its value). Option (4) is still the only option that describes the correct change, so the question can still be solved.