A certain metallic surface is illuminated by monochromatic radiation of wavelength \lambda . The stopping potential for photoelectric current for this radiation is 3V0. If the same surface is illuminated with a radiation of wavelength 2 \lambda , the stopping potential is V0. The threshold wavelength of this surface for photoelectric effect is \underline\hspace2cm\lambda .

Dual Nature of Radiation PYQ — JEE Main 2021 · 20 Jul · Shift 2 · Q71 with Solution

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Correct answer: 4

For incident radiation of wavelength $\lambda$ ,

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

where $\phi$ is the work function and $V_s$ is the stopping potential.

\frac{hc}{\lambda}=\phi + 3eV_0 \qquad (1)

\frac{hc}{2\lambda}=\phi + eV_0 \qquad (2)

\frac{hc}{\lambda}-\frac{hc}{2\lambda}=3eV_0-eV_0

\frac{hc}{2\lambda}=2eV_0

eV_0=\frac{hc}{4\lambda}

From equation (2):

\phi = \frac{hc}{2\lambda}-eV_0

Substitute $eV_0=\dfrac{hc}{4\lambda}$ :

\phi = \frac{hc}{2\lambda}-\frac{hc}{4\lambda} =\frac{hc}{4\lambda}

At threshold,

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

So,

\frac{hc}{\lambda_0}=\frac{hc}{4\lambda}

\lambda_0=4\lambda

Hence, the blank is filled by:

4