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A metal plaet of area $1\times 10^{-4 }m^{2}$ is iluminated by a radiation of intensity 16 mW/ m² . The work function of the metal is 5eV . The energy of the incident photons is 10 eV and only 10% of it produces photo electrons.The number of emitted photo electron per second and their maximum energy, respectively,will be:

$\left [1eV= 1.6\times 10^{-19 }J \right ]$

Option 1)
$10^{10}\: and \: 5eV$
Option 2)
$10^{12}\: and \: 5eV$
Option 3)
$10^{11}\: and \: 5eV$
Option 4)
$10^{14}\: and \: 10eV$

Answers (1)

best_answer

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

Energy of a photon -

$\fn_jvn E= h\nu = \frac{hc}{\lambda }$

- wherein

$h= Plank's\: constant$

$\boldsymbol{\nu= frequency\: of \: radiation }$

$\lambda \rightarrow wave \: length$

$KE_{max} = E - \phi$

$=10 eV - 5eV = 5eV$

$\frac{n}{t} = \frac{IA}{E} = \frac{16\times 10^{-3}\times 10^{-4}}{10\times 1.6\times 10^{-18}} = 10^{11}$

Option 1)

$10^{10}\: and \: 5eV$

Option 2)

$10^{12}\: and \: 5eV$

Option 3)
$10^{11}\: and \: 5eV$
Option 4)

$10^{14}\: and \: 10eV$

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