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Radiation of wavelength λ, is incident on a photocell. The fastest emitted electron has speed $v$ . If the wavelength is changed
to $\frac{3\lambda }{4}$ the speed of the fastest emitted electron will be :

Option 1)
$> v\left ( \frac{4}{3} \right )^{\frac{1}{2}}$

Option 2)
$< v\left ( \frac{4}{3} \right )^{\frac{1}{2}}$

Option 3)
$= v\left ( \frac{4}{3} \right )^{\frac{1}{2}}$

Option 4)
$= v\left ( \frac{3}{4} \right )^{\frac{1}{2}}$

Answers (1)

best_answer

As we have learnt,

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

From Einstein equation,

$\\\frac{1}{2}mv^2 = \frac{hc}{\lambda}- \phi\;\;\;-(1) \\\frac{1}{2}mv'^2 = \frac{hc}{\frac{3\lambda}{4}}- \phi\;\;\;-(2)$

$\\\frac{1}{2}mv'^2 =\frac{4}{3} \left(\frac{hc}{\lambda}- \phi\right)+\frac{\phi}{3}$

$=\frac{4}{3} \left(\frac{1}{2}mv^2\right)+{\phi}$

$\\v'^2 = \frac{4}{3}v^2 + \frac{2\phi}{m}\\v' = \sqrt{\frac{4}{3}v^2 + \frac{2\phi}{m}}$

$\therefore v' > v\cdot \left(\frac{4}{3}\right)^{\frac{1}{2}}$

Option 1)

$> v\left ( \frac{4}{3} \right )^{\frac{1}{2}}$

Option 2)

$< v\left ( \frac{4}{3} \right )^{\frac{1}{2}}$

Option 3)

$= v\left ( \frac{4}{3} \right )^{\frac{1}{2}}$

Option 4)

$= v\left ( \frac{3}{4} \right )^{\frac{1}{2}}$

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Avinash

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