Question
In astronomical observations, signals observed from the distant starts are generally weak. If the photon detector receives a total of $=3.15 \times 10^{-18} J$ from the radiations of 600 nm . Calculate the number of photons received by the detector.

Answer

Total energy received $=3.15 \times 10^{-18}$
$
\lambda=600 nm=600 \times 10^{-9} m=6 \times 10^{-7} m
$
The energy of one photon, $E =\frac{h c}{\lambda}=\frac{\left(6.626 \times 10^{-34} Js \right) \times\left(3 \times 10^8 ms^{-1}\right)}{\left(6 \times 10^{-7} m\right)}$
$
=3.3125 \times 10^{-19} J
$
$\therefore$ No. of photons $=\frac{3.15 \times 10^{-18} J}{3.3125 \times 10^{-19} J}=10$

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