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A photoelectric surface is illuminated successively by monochromatic light of wavelength $\lambda$ and $\frac{\lambda } {2}$ . If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface of the material is :

(h = Planck's constant, c = speed of light)

Option 1)
$\frac{{\text{hc}}} {\lambda }$

Option 2)
$\frac{{\text{2 hc}}} {\lambda }$

Option 3)
$\frac{{\text{hc}}} {{3\lambda }}$

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

Answers (1)

best_answer

As discussed in

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

Let work function be $\phi$

$K_{1}=K.E_{max}=\frac{hc}{\lambda}-\phi \:\:\:\:\:\:\:\:\:\:\:-(i)$

$K_{2}=\frac{hc}{\frac{\pi}{2}}-\phi=2 \frac{hc}{\lambda}-\phi \:\:\:\:\:\:\:\:\:\:\:-(ii)$

According to the question,

$\frac{2hc}{\lambda}-\phi=3 \left( \frac{hc}{\lambda}- \phi \right )$

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

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

Or, $\phi=\frac{hc}{2\lambda}$

Option 1)

$\frac{{\text{hc}}} {\lambda }$

This option is incorrect.

Option 2)

$\frac{{\text{2 hc}}} {\lambda }$

This option is incorrect.

Option 3)

$\frac{{\text{hc}}} {{3\lambda }}$

This option is incorrect.

Option 4)

$\frac{{\text{hc}}} {{2\lambda }}$

This option is correct.

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Plabita

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