Question

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

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

Answer 1

Answer: (A)

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