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Kelvi

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A certain metallic surface is illuminated with monochromatic light of wavelength, $\lambda.$ The stopping potential for photoelectric current for this light is $3V_0.$ If the same surface is illuminated with light of wavelength $2\lambda,$ the stopping potential is $V_0.$ The threshold wavelength for this surface for photoelectric effect is

hard

Dual Nature of Radiation and Matter

2015

physics

$4\lambda$

$\frac{\lambda}{4}$

$\frac{\lambda}{6}$

Explanation

To solve this problem, we will use the photoelectric effect equation: $\\ E = h \nu = \frac{hc}{\lambda} \\$ The stopping potential $V_s$

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$6\lambda$

is related to the maximum kinetic energy of the emitted electrons:

\\ eV_s = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} \\

where

\lambda_0

is the threshold wavelength.

\\

Given:

\\

• For wavelength

\lambda,

the stopping potential is

3V_0: \\ e \cdot 3V_0 = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} \quad (1) \\

• For wavelength

2\lambda,

the stopping potential is

V_0: \\ e \cdot V_0 = \frac{hc}{2\lambda} - \frac{hc}{\lambda_0} \quad (2) \\

Subtract equation (2) from equation (1):

\\ e \cdot 3V_0 - e \cdot V_0 = \left(\frac{hc}{\lambda} - \frac{hc}{2\lambda}\right) \\ 2eV_0 = \frac{hc}{\lambda} - \frac{hc}{2\lambda} \\ 2eV_0 = \frac{hc}{\lambda} \left(1 - \frac{1}{2}\right) \\ 2eV_0 = \frac{hc}{2\lambda} \\

From this, we can express

\frac{hc}{2\lambda}

in terms of

V_0: \\ \frac{hc}{2\lambda} = 2eV_0 \\

Substitute

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

into equation (2):

\\ eV_0 = 2eV_0 - \frac{hc}{\lambda_0} \\ \frac{hc}{\lambda_0} = 2eV_0 - eV_0 \\ \frac{hc}{\lambda_0} = eV_0 \\

Thus, the threshold wavelength

\lambda_0

is:

\\ \lambda_0 = \frac{hc}{eV_0} \\

Since

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

we have:

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

Therefore,

\lambda_0 = 4\lambda. \\

This corresponds to Option 1.