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

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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