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A certain metallic surface is illuminated with monochromatic light of wavelength, $\lambda$ . The stopping potential for photo-electric current for this light is 3V₀. If the same surface is illuminated with light of wavelength $2\lambda$ , the stopping potential is V₀. The threshold wavelength for this surface for photo-electric effect is:

Option 1)
$4 \lambda$

Option 2)
$\frac{\lambda}{4}$

Option 3)
$\frac{\lambda}{6}$

Option 4)
$6 \lambda$

Answers (1)

best_answer

Stopping Potential /Cut-off Potential -

It is defined as the potential necessary to stop any electron from reaching the other side.

-

and

Conservation of energy -

$h\nu = \phi _{0}+\frac{1}{2}mv^{2}_{max}$

$h\nu = h\nu _{0}+\frac{1}{2}mv^{2}_{max}$

$h\left ( \nu-\nu _{0} \right )=\frac{1}{2}mv^{2}_{max}$

$where, h - Plank's constant\ \nu - Frequency\ \nu_{0} - threshold\ frequency\ \phi_{0}- work function$

- wherein

1st Case: $3eV_{0}= \frac{hc}{\lambda } - \phi - (1)$

2nd Case: $eV_{0} = \frac{hc}{2\lambda } - \phi - (2) \times 3$

$\Rightarrow 3eV_{0} = \frac{3hc}{2\lambda } - 3\phi - (3)$

From equation (1) and (3)

$\frac{3hc}{2\lambda } - 3\phi = \frac{hc}{\lambda } - \phi$

or $2\phi = \frac{hc}{2\lambda }.$

or $\phi = \frac{hc}{4\lambda }$

if threshold wavelength is $\lambda _{0}$ then

$\frac{hc}{4\lambda }$

or $\lambda _{0} = 4\lambda$

Option 1)

$4 \lambda$

Option 2)

$\frac{\lambda}{4}$

Option 3)

$\frac{\lambda}{6}$

Option 4)

$6 \lambda$

Posted by

prateek

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