Question

An electron, accelerated by a potential difference V, has de-Broglie wavelength $ \lambda $. If the electron is accelerated by a potential difference 4V, its de-Broglie wavelength will be

• A. 2 $ \lambda $
• B. 4$ \lambda $
• C. $ \frac{ \lambda }{ 2} $
• D. $ \frac{ \lambda }{ 4} $

Answer 1

Answer: (C)

From the relation,
$ \lambda = \frac{ h }{ \sqrt{ 2 m e V }} $ ..(i)
So, when V becomes 4V, then
$ \lambda ' = \frac{ h }{ \sqrt{ 2 m e (4 V) }} $
$ \Rightarrow \lambda ' = \frac{ h }{ 2 \sqrt{ 2 m e V }} $
$ \Rightarrow \lambda ' = \frac{ \lambda }{ 2} $