Question

An electron accelerated through a potential of $10,000 \,V$ from rest has a de-Broglie wave length $‘\lambda’$. What should be the accelerating potential so that the wave length is doubled ?

• A. 20,000 V
• B. 40,000 V
• C. 5,000 V
• D. 2.500 V

Answer 1

Answer: (D)

$\because$ Kinetic energy of a electron due to accelerated by a potential $V, KE =e V$
$ \frac{1}{2} m_{e} v^{2} =e V $
$\Rightarrow \frac{1 \times p^{2}}{2 m_{e}} =e V$
$[\because p=m v]$
$\therefore p=\sqrt{2 e V m_{e}}$
$\because$ de-Broglie wavelength of a particle having momentum $p$,
$\lambda=\frac{h}{p}$
According to question,
$ \frac{\lambda_{1}}{\lambda_{2}}=\frac{p_{2}}{p_{1}}=\frac{\sqrt{2 \,e V_{2} m_{e}}}{\sqrt{2 \,e V_{1} m_{e}}}=\sqrt{\frac{V_{2}}{V_{1}}}$
$\because \lambda_{2}=2 \lambda_{1}$ (given)
$\Rightarrow \frac{\lambda_{1}}{2 \lambda_{1}}=\sqrt{\frac{V_{2}}{10000}}$
$\therefore $ Potential $V_{2}=\frac{10^{4}}{4}=2500\, V$