The beam of light has three wavelengths 4144 , A, 4972 ; A and 62… - Vidyadip

MCQ

The beam of light has three wavelengths $4144 \,\mathring A$, $4972 \;\mathring A$ and $6216\; \mathring A$ with a total intensity of $3.6 \times$ $10^{-5}\,Wm ^2$ equally distributed amongst the three wavelengths. The beam falls normally on the area $1\,cm ^2$ of a clean metallic surface of work function $2.3\,eV$. Assume that there is no loss of light by reflection and that each energetically capable photon ejects one electron. Calculate the number of photoelectrons liberated in $2\,s$.

A
$2 \times 10^9$
✓
$1.075 \times 10^{12}$
C
$9 \times 10^8$
D
$3.75 \times 10^6$

✓

Answer

Correct option: B.

$1.075 \times 10^{12}$

b
(b)

$\left(\lambda_0\right)=\frac{ hc }{\phi}=5404\,A$

For each wavelength energy incident on the surface per unit time

= intensity of each $\times$ area of the surface wavelength $=1.2 \times 10^{-7}$ joule

$E=\left(1.2 \times 10^{-7}\right) \times 2=2.4 \times 10^{-7}\,J$

Number of photons $n _1$ due to wavelength $4144\,A$

$n_1=\frac{\left(2.4 \times 10^{-7}\right)\left(4144 \times 10^{-10}\right)}{\left(6.63 \times 10^{-34}\right)\left(3 \times 10^8\right)}=0.5 \times 10^{12}$

Number of photons $n_2$ due to the wavelength $4972 \mathring A$

$n _2=\frac{\left(2.4 \times 10^{-7}\right)\left(4972 \times 10^{-10}\right)}{\left(6.63 \times 10^{-34}\right)\left(3 \times 10^3\right)}=0.572 \times 10^{12} $

$N = n _1+ n _2=1.075 \times 10^{12}$

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