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

Answer
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Hint: The number of photons can be found by the ratio of the energy given to the energy obtained from the wavelength given. Using this relation try to find out the number of photons that are received by the detector if the radiation of the photon is detected with a wavelength of \[600nm\]

Complete step by step answer:
 Given that the total energy received by the photon detected is \[E=3.15\times {{10}^{-18}}J\]
The wavelength of the detected photon is obtained as \[\lambda =600nm=600\times {{10}^{-9}}m\]
We know that the energy of one photon is \[=h\nu =h\dfrac{c}{\lambda }\]
Where
\[h\]= Is the Planck’s constant whose value is $6.626\times {{10}^{-34}}Js$
\[\nu \]= frequency of the photon (whose units are ${{m}^{-1}}$ )
\[c\]= speed of the light in the vacuum whose value is $3\times {{10}^{8}}m/s$
\[\lambda \]= wavelength of the detected photon (the units of wavelength are $m$ )
 Put all these values in the above energy formula and calculate the energy of the detected photon
After substituting the above values we obtain
\[E=h\dfrac{c}{\lambda }=6.626\times {{10}^{-34}}Js\times \dfrac{3\times {{10}^{8}}m/s}{600\times {{10}^{-9}}m}=3.313\times {{10}^{-19}}J\]
This is the energy of the one photon.
Given the energy of the total number of photons is \[E=3.15\times {{10}^{-18}}J\]
So the total number of photons is \[\dfrac{3.15\times {{10}^{-18}}}{3.313\times {{10}^{-19}}}=9.51\tilde{\ }10\]

Therefore the total number of the photons received by the detector is equal to $10$ .

Note: While doing the calculations make sure that all the values of the terms involved in the formula should be in the same system of units. If they are not in the same system of units then first convert the units in such a way that all the terms are in the same system of units. Also the units should balance on both sides of the equation or the formula; because of this reason in the above solution we converted the units of the wave length from the nanometers to the meters. One nanometer is equal to ${{10}^{-9}}$ meters.