Electron beam used in an electron microscope, when accelerated by a voltage of 20 kV, has a de-Broglie wavelength of \lambda_0 . IF the voltage is increased to 40 kV, then the de-Broglie wavelength associated with the electron beam would be :

Dual Nature of Radiation PYQ — JEE Main 2023 · 25 Jan · Shift 1 · Q56 with Solution

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Correct answer: C

For an electron accelerated through a potential difference $V$ , its de-Broglie wavelength is

\lambda = \frac{h}{p}

For non-relativistic electrons,

\frac{p^2}{2m} = eV

So,

p = \sqrt{2meV}

Hence,

\lambda = \frac{h}{\sqrt{2meV}}

This shows that

\lambda \propto \frac{1}{\sqrt{V}}

When the voltage is increased to

V_2 = 40\,\text{kV}

we use

\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{V_2}}

Thus,

\frac{\lambda_2}{\lambda_0} = \sqrt{\frac{20}{40}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}

Therefore,

\lambda_2 = \frac{\lambda_0}{\sqrt{2}}

So the correct option is C.