Question

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

• A. $0.451\text{nm}$
• B. $0.361\text{nm}$
• C. $0.153\text{nm}$
• D. $0.281\text{nm}$

Answer 1

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

de Broglie wavelength associated with electron is,

$\lambda =\frac{h}{p}=\frac{h}{\sqrt{2m_{e}eV}}$

For electron, putting the values, we get,

$\lambda =\frac{1.227}{\sqrt{V}}\text{nm}$

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