Question

An electron accelerated through a potential of $10,000V$ 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 $V,KE=eV$
$\frac{1}{2}{m}_e{v}^2=eV$
$\Rightarrow \frac{1\times p^2}{2m_e}=eV$
$[\because p=mv]$
$\therefore p=\sqrt{2eV{m}_e}$
$\because$ de-Broglie wavelength of a particle having momentum $p$,
$\lambda =\frac{h}{p}$
According to question,
$\frac{{\lambda }_1}{{\lambda }_2}=\frac{p_2}{p_1}=\frac{\sqrt{2e{V}_2{m}_e}}{\sqrt{2e{V}_1{m}_e}}=\sqrt{\frac{V_2}{V_1}}$
$\because {\lambda }_2=2{\lambda }_1$ (given)
$\Rightarrow \frac{{\lambda }_1}{2{\lambda }_1}=\sqrt{\frac{V_2}{10000}}$
$\therefore$ Potential $V_2=\frac{10^4}{4}=2500V$