Question

Calculate the
(i) momentumand
(ii) de-Broglie wavelength of electrons accelerated through a potential difference of $56V$.

• A. $p=4.02\times 10^{-24}kgm{s}^{-1},\lambda =0.164nm$.
• B. $p=2.5\times 10^{22}kgm{s}^{-1},\lambda =1nm$
• C. $p=1.25\times 10^{-22}kgm{s}^{-1},\lambda =0.5nm$
• D. None of the above

Answer 1

Answer: (A)

Given, potential difference, $V=56V$
(i) Use the formula for kinetic energy
$eV=\frac{1}{2}m{v}^2$
$\Rightarrow \frac{2eV}{m}=v^2$
$\Rightarrow v=\sqrt{\frac{2eV}{m}}$
where, $m$ is mass and $v$ is velocity of electron.
Momentum associated with accelerated electron,
$p=mv=m\sqrt{\frac{2eV}{m}}=\sqrt{2eVm}$
$=\sqrt{2\times 1.6\times 10^{-19}\times 56\times 9\times 10^{-31}}$
$=4.02\times 10^{-24}kg-m{s}^{-1}$
(ii) de-Broglie wavelength of electron,
$\lambda =\frac{12.27}{\sqrt{V}}\mathring{A}=\frac{12.27}{\sqrt{56}}$
$=0.164\times 10^{-9}m=0.164nm$