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A beam of light has two wave lengths 4972 Å and 6216 Å with a total intensity of 3.6 $\times$ 10^-3 Wm^-2 equally distributed
among the two wavelengths. The beam falls normally on an area of 1 cm² of a clean metallic surface of work function 2.3 eV.
Assume that there is no loss of light by reflection and that each capable photon ejects one electron. The number of photo
electrons liberated in 2s is approximately :

Option 1)
$6\times 10^{11}$

Option 2)
$9\times 10^{11}$

Option 3)
$11\times 10^{11}$

Option 4)
$15\times 10^{11}$

Answers (1)

best_answer

As we have learned

Energy of a photon -

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

- wherein

$h= Plank's\: constant$

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

$\lambda \rightarrow wave \: length$

$\lambda _1 = 4972 A \degree , \lambda _2 = 6216 A \degree \\ I = 3.6 \times 10^{-3} w/m^2$

Intensity with each wave length = $1.8 \times 10^{-3}W / m^2$

$\phi = \frac{hc}{\lambda }= \frac{6.62 \times 10^{-34}\times 3\times 10^8}{\lambda }$

$\phi = 12.4 \times 10^3/\lambda$

$\phi _1 = 12.4 \times 10^3/\lambda _1= 2.493 eV = 3.98 \times 10^{-19}J$

$\phi _2 = 12.4 \times 10^3/\lambda _2=3.189 \times 10^{-19} J (1.99eV )$

$\phi =2.3 eV$

$\phi_2 < \phi _1 \\ N /sec = \frac{1.8\times 10^{-3}}{3.984\times 10^{-19}\times 10^{-4}}= 0.45 \times 10^{12}$

N = $9 \times 10^{11}$

Option 1)

$6\times 10^{11}$

Option 2)

$9\times 10^{11}$

Option 3)

$11\times 10^{11}$

Option 4)

$15\times 10^{11}$

Posted by

Avinash

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