Question: Q. 5. Monochromatic light of frequency $6.0 \times 10^{14} \mathrm{~Hz}$ is produced by a laser. The power emitted is $2.0 \times$ $10^{-3} \mathrm{~W}$. Calculate the (i) energy of a photon in the light beam and (ii) number of photons emitted on an average by the source. [CBSE Comptt. 2018]
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Solution:
Ans. Calculating
(i) Energy of a photon
$1 / 2+1 / 2$
(ii) Number of photons emitted
Energy of photon $=h \mathrm{v}$
$=6.63 \times 10^{-34} \times 6.0 \times 10^{14} \mathrm{~J}$
$=3.978 \times 10^{-19} \mathrm{~J}$
$\cong 2.49 \mathrm{eV}$ Number of photons emitted per second Power
$=\overline{\text { Energy of photon }}$
$=\frac{2.0 \times 10^{-3} \mathrm{~J} / \mathrm{s}}{3.978 \times 10^{-19} \mathrm{~J}}$
$=5.03 \times 10^{15}$ photons $/$ second
[CBSE Marking Scheme 2018]
