The kinetic energy of emitted electron is E when the light incident on the metal has wavelength \lambda . To double the kinetic energy, the incident light must have wavelength:

Dual Nature of Radiation PYQ — JEE Main 2022 · 29 Jul · Shift 1 · Q63 with Solution

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Correct answer: B

For incident light of wavelength $\lambda$ , photon energy is

\frac{hc}{\lambda}

and the maximum kinetic energy of emitted electron is given as $E$ .

So,

E = \frac{hc}{\lambda} - \phi

where $\phi$ is the work function of the metal.

We want the new kinetic energy to be $2E$ for some new wavelength $\lambda'$ .

Then,

2E = \frac{hc}{\lambda'} - \phi

From the first equation,

\phi = \frac{hc}{\lambda} - E

Substitute into the second equation:

2E = \frac{hc}{\lambda'} - \left(\frac{hc}{\lambda} - E\right)

Simplify:

2E = \frac{hc}{\lambda'} - \frac{hc}{\lambda} + E

E = \frac{hc}{\lambda'} - \frac{hc}{\lambda}

So,

\frac{hc}{\lambda'} = E + \frac{hc}{\lambda}

Taking reciprocal form,

\lambda' = \frac{hc}{E + \frac{hc}{\lambda}}

Now simplify:

\lambda' = \frac{hc}{\frac{E\lambda + hc}{\lambda}} = \frac{hc\lambda}{E\lambda + hc}

Thus the required wavelength is

\boxed{\frac{hc\lambda}{E\lambda + hc}}

This matches Option B.