Question

The de-Broglie wavelength associated with an electron accelerated through a potential difference of $64\text{V}$ is,

• A. $1.227\text{nm}$
• B. $0.613\text{nm}$
• C. $0.302\text{nm}$
• D. $0.153\text{nm}$

Answer 1

The correct option is D $0.153\text{nm}$

When an electron is accelerated through a potential difference $V$, the de-Broglie wavelength is given by,

$\lambda =\frac{1.227}{\sqrt{V}}\text{(here,}\lambda \text{is in nm)}$

Now, $V=64\text{volts}$

$\Rightarrow \lambda =\frac{1.227}{\sqrt{64}}=\frac{1.227}{8}=0.153\text{nm}$

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