When radiation of wavelength \lambda is incident on a metallic surface, the stopping potential of ejected photoelectrons is 4.8 V. If the same surface is illuminated by radiation of double the previous wavelength, then the stopping potential becomes 1.6 V. The threshold wavelength of the metal is :

Dual Nature of Radiation PYQ — JEE Main 2021 · 25 Jul · Shift 2 · Q62 with Solution

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Correct answer: B

For incident wavelength $\lambda$ ,

\frac{hc}{\lambda} = \phi + eV_{s1}

where $V_{s1}=4.8\,\text{V}$ .

For incident wavelength $2\lambda$ ,

\frac{hc}{2\lambda} = \phi + eV_{s2}

where $V_{s2}=1.6\,\text{V}$ .

Since $1\,\text{eV} = e\times 1\,\text{V}$ , we can write energies directly in eV.

So,

\frac{hc}{\lambda} = \phi + 4.8

\frac{hc}{2\lambda} = \phi + 1.6

E = \frac{hc}{\lambda}

Then,

E = \phi + 4.8

\frac{E}{2} = \phi + 1.6

From

E - \frac{E}{2} = (\phi+4.8) - (\phi+1.6)

we get

\frac{E}{2} = 3.2

E = 6.4\,\text{eV}

Using

\phi = E - 4.8 = 6.4 - 4.8 = 1.6\,\text{eV}

At threshold,

\phi = \frac{hc}{\lambda_0}

Also,

E = \frac{hc}{\lambda} = 6.4\,\text{eV}

Thus,

\frac{E}{\phi} = \frac{hc/\lambda}{hc/\lambda_0} = \frac{\lambda_0}{\lambda}

So,

\frac{\lambda_0}{\lambda} = \frac{6.4}{1.6} = 4

Hence,

\lambda_0 = 4\lambda

The threshold wavelength is $4\lambda$ .

So the correct option is B.