Question

Electrons used in an electron microscope are accelerated by a voltage of $25\, kV$. If the voltage is increased to $100\, kV$ then the de-Broglie wavelength associated with the electrons would

• A. increase by 2 times
• B. decrease by 2 times
• C. decrease by 4 times
• D. increase by 4 times

Answer 1

The Correct Option is B

$\lambda=\frac{ h }{\sqrt{2 \,mq \,\Delta V }} \propto \frac{1}{\sqrt{\Delta V }}$
$\frac{\lambda_{2}}{\lambda_{1}}=\sqrt{\frac{\Delta V _{1}}{\Delta V _{2}}}=\sqrt{\frac{25}{100}}=\frac{1}{2}$
$\lambda_{2}=\frac{\lambda_{1}}{2}$

Answer 2

De Broglie wavelength in terms of accelerated potential difference is

 λ= \(\frac{12.27}{\sqrt{V}}\)

​The initial and final wavelengths are λ1and λ2 having potential as V1 and V2

​\(\frac{\lambda_1}{\lambda_2} = \sqrt\frac{V_1}{V_2}\)

 ​\(\frac{\lambda_1}{\lambda_2} = \sqrt \frac{100}{25}\)

 ​\(\frac{\lambda_1}{\lambda_2} = \sqrt4\)

 \(\frac{\lambda_1}{\lambda_2} = 2\)

 \(\lambda_2 = \frac{\lambda_1}{2}\)

 So, the new wavelength is decreased by 2.

The correct option is (B)