Question.
In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of $3.15 \times 10^{-18} \mathrm{~J}$ from the radiations of $600 \mathrm{~nm}$, calculate the number of photons received by the detector.
In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of $3.15 \times 10^{-18} \mathrm{~J}$ from the radiations of $600 \mathrm{~nm}$, calculate the number of photons received by the detector.
Solution:
From the expression of energy of one photon $(E)$,
$E=\frac{\text { he }}{\lambda}$
Where, $\lambda=$ wavelength of
radiation $h=$ Planck's
constant $c=$ velocity of
radiation
Substituting the values in the given expression of $E$ :
$=\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(3 \times 10^{8} \mathrm{~ms}^{-1}\right)}{\left(600 \times 10^{-9} \mathrm{~m}\right)}=3.313 \times 10^{-19} \mathrm{~J}$
Energy of one photon $=3.313 \times 10^{-19} \mathrm{~J}$
Number of photons received with $3.15 \times 10^{-18} \mathrm{~J}$ energy
$=\frac{3.15 \times 10^{-18} \mathrm{~J}}{3.313 \times 10^{-19} \mathrm{~J}}$
$=9.5$
$\approx 10$
From the expression of energy of one photon $(E)$,
$E=\frac{\text { he }}{\lambda}$
Where, $\lambda=$ wavelength of
radiation $h=$ Planck's
constant $c=$ velocity of
radiation
Substituting the values in the given expression of $E$ :
$=\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(3 \times 10^{8} \mathrm{~ms}^{-1}\right)}{\left(600 \times 10^{-9} \mathrm{~m}\right)}=3.313 \times 10^{-19} \mathrm{~J}$
Energy of one photon $=3.313 \times 10^{-19} \mathrm{~J}$
Number of photons received with $3.15 \times 10^{-18} \mathrm{~J}$ energy
$=\frac{3.15 \times 10^{-18} \mathrm{~J}}{3.313 \times 10^{-19} \mathrm{~J}}$
$=9.5$
$\approx 10$