Question

The kinetic energy of an electron is $5\text{eV}$. Calculate the de-Broglie wavelength associated with it. $(h=6.6\times {10}^{-34}\text{Js},m_e=9.1\times {10}^{-31}\text{kg})$

• A. $5.47\mathring{\text{A}}$
• B. $10.9\mathring{\text{A}}$
• C. $2.7\mathring{\text{A}}$
• D. $\text{None of these}$

Answer 1

The correct option is A $5.47\mathring{\text{A}}$

de Broglie wavelength is given by,
$\lambda =\frac{h}{\sqrt{2m_e{E}_k}}$

Here, $E_k$ is kinetic energy and $m_e$ is mass of the electron.
$\lambda =\frac{6.6\times {10}^{-34}}{\sqrt{2\times 9.1\times {10}^{-31}\times 5\times 1.6\times {10}^{-19}}}$

$\Rightarrow \lambda =5.469\times {10}^{-10}\text{m}\approx 5.47\mathring{\text{A}}$

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