The de-Broglie wavelength associated with an electron accelerated through a potential difference V is λ. The new de Broglie wavelength of electron when potential difference is increased to 9V will be:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{\lambda}{3} \)
The de Broglie wavelength is \( \lambda = h/p \). An electron accelerated from rest through a potential difference V gains kinetic energy \( eV \), so its momentum is \( p = \sqrt{2meV} \).
\( \displaystyle \lambda = \frac{h}{\sqrt{2meV}} \Rightarrow \lambda\propto\frac{1}{\sqrt{V}} \)
\( \displaystyle \frac{\lambda'}{\lambda} = \sqrt{\frac{V}{9V}} = \frac{1}{3} \Rightarrow \lambda' = \frac{\lambda}{3} \)