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

Asked by Abhisek | 1 year ago |  67

1 Answer

Solution :-

From the expression of energy of one photon (E),

\(E=\dfrac{hc}{\lambda }\)

Where,

λ denotes the wavelength of the radiation

h is Planck’s constant

c denotes the velocity of the radiation

Substituting these values in the expression for E:

\(E=\dfrac{(6.626\times 10^{-34}Js)(3\times 10^{8}ms^{-1})}{(600\times 10^{-9})}\)

\( =3.313\times 10^{-19}\, J\)

Energy held by one photon 

=\(3.313\times 10^{-19}\, J\)

No. photons received with \(3.15\times 10^{-18}J\;energy\)

\(=\dfrac{3.15\times 10^{-18}J}{3.313\times 10^{-19}J }\\ \\ \)

\( =9.5\\ \\ \approx 10\)

Answered by Pragya Singh | 1 year ago

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