Question

What will be the final linear momentum of an electron, when it is accelerated through a potential difference of $100\text{V}$, if it is initially at rest?

• A. $9.1\times {10}^{-24}\text{kg m/s}$
• B. $6.5\times {10}^{-24}\text{kg m/s}$
• C. $5.4\times {10}^{-24}\text{kg m/s}$
• D. $1.6\times {10}^{-24}\text{kg m/s}$

Answer 1

The correct option is C $5.4\times {10}^{-24}\text{kg m/s}$

The de-Broglie wavelength of an electron is,
$\lambda =\frac{h}{p}=\frac{h}{\sqrt{2m(K.E)}}=\frac{h}{\sqrt{2meV}}$
where $m$ is the mass of an electron and $e$ is the charge of an electron.
Substituting the numerical values of $h,m$ and $e$ we get
$\lambda =\frac{12.27}{\sqrt{V}}\left(\text{in}\mathring{\text{A}}\right)$

where $V$ is the magnitude of accelerating potential difference, in volts.

For a $100\text{V}$ accelerating potential difference,

$\lambda =\frac{12.27}{\sqrt{100}}\mathring{\text{A}}=1.227\mathring{\text{A}}$
Now, $\lambda =\frac{h}{p}\Rightarrow p=\frac{h}{\lambda }$
$\therefore p=\frac{6.63\times {10}^{-34}}{1.227\times {10}^{-10}}=5.4\times {10}^{-24}\text{kg m/s}$

Hence, $\left(C\right)$ is the correct answer.