Question

The linear momentum of an electron, initially at rest, accelerated through a potential difference of $ 100 \,V $ is

• A. $ 9.1 \times10^{-24} $
• B. $ 6.5 \times10^{-24} $
• C. $ 5.4 \times10^{-24} $
• D. $ 1.6 \times10^{-24} $

Answer 1

Answer: (C)

de-Broglie wavelength of electron is given by
$\lambda=\frac{h}{mv}=\frac{h}{\sqrt{2mE}}$
Substituting the value of $E$, we get
$\lambda=\frac{h}{\sqrt{2meV}}$
Here $m=9.1\times10^{-31} kg; e=1.6\times10^{-19}C$
and $h=6.6\times10^{-34}Js$
we get $ \lambda=\frac{12.27}{\sqrt{V}}\times10^{-10}=\frac{12.27}{\sqrt{V}}\mathring{A}$
The de-Broglie wavelength of electrons, when accelerated through a potential difference of $100\, V$ will be
$\lambda=\frac{12.27}{\sqrt{100}}=1.227\mathring{A}$
Moreover, $\lambda=\frac{h}{p}$
$\Rightarrow p=\frac{6.6\times10^{-34}}{1.227\times10^{-10}}$
$=5.5\times10^{-24} kg -ms^{-1}$