The de-Broglie wavelength associated with an electron accelerated through a potential difference $V$ is $\lambda$. What will be its wavelength when the accelerating potential is increased to $4V$?

CBSE Class 12 · 2006 · All India · Modern PhysicsVery Short Answer

The de-Broglie wavelength associated with an electron accelerated through a potential difference

V

λ

. What will be its wavelength when the accelerating potential is increased to

4 V

Answer: $λ /2$

The de-Broglie wavelength of an accelerated charge is

λ = \frac{h}{2 m e V} \Rightarrow λ \propto \frac{1}{V} .

Hence

\frac{λ ^{'}}{λ} = \frac{V}{4 V} = \frac{1}{2}

λ^{'} = \frac{λ}{2}

CBSE Class 12 Board · 2006 · All India · Subjective