QJEEM23Ph9a11: JEE Main 2023 Physics - Class 12 JEEM23Ph9a11 Solution: The de Broglie wavelength of an electron having kinetic energy…

$\lambda = \frac{h}{\sqrt{2mE}}$

where $h$ is Planck's constant, $m$ is the mass of the particle, and $E$ is its kinetic energy.

We are given that the de Broglie wavelength of an electron with kinetic energy $E$ is $\lambda$ , and we want to find the de Broglie wavelength of the same electron when its kinetic energy becomes $\frac{E}{4}$ .

To do this, we can use the formula for the de Broglie wavelength again, but with the new kinetic energy $\frac{E}{4}$ :

$\lambda' = \frac{h}{\sqrt{2m\left(\frac{E}{4}\right)}} = \frac{2h}{\sqrt{2mE}} = 2\lambda$

where we have used the fact that $\sqrt{\frac{1}{4}} = \frac{1}{\sqrt{4}} = \frac{1}{2}$ to simplify the expression.

Therefore, the de Broglie wavelength of the electron when its kinetic energy becomes $\frac{E}{4}$ is twice its original value, or $\boxed{2\lambda}$ .