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A photoelectric surface is illuminated successively by monochromatic light of wavelengths $\lambda$ 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 is :

Option 1)
$\frac{hc}{3\lambda }$

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

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

Option 4)
$\frac{3hc}{\lambda }$

Answers (1)

best_answer

As we have learned

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

$1/2 mv^2 = K.E = \frac{hc }{\lambda }- \phi$

$K_1 = \frac{hc}{\lambda }- \phi$

$K_2 = \frac{hc}{\lambda /2 }- \phi = 2 \frac{hc}{\lambda }- \phi \\$

$K_2 = 3K_1 \\\Rightarrow 2\frac{hc}{\lambda } - \phi = 3 \frac{hc}{\lambda } - 3\phi$

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

Option 1)

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

Option 2)

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

Option 3)

$\frac{hc}{\lambda }$

Option 4)

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

Posted by

Avinash

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