A star which can be seen with naked eye from Earth has intensity $1.6 \times 10^{-9} \mathrm{Wm}^{-2}\( on Earth. If the corresponding wavelength is \)560 \mathrm{~nm}\(, and the radius of the human eye is \)2.5 \times 10^{-3} \mathrm{~m}\(, the number of photons entering in our in \)1 \mathrm{~s}$ is..... (A) \(7.8 \times 10^{4} \mathrm{~s}^{-1}\) (B) \(8.85 \times 10^{4} \mathrm{~s}^{-1}\) (C) \(7.85 \times 10^{5} \mathrm{~s}^{-1}\) (D) \(8.85 \times 10^{5} \mathrm{~s}^{-1}\)

To calculate the energy of a single photon, we will use the equation: \(E = \dfrac{hc}{\lambda}\) where \(E\) is the energy of the photon, \(h\) is Planck's constant (\(6.626 \times 10^{-34} \mathrm{Js}\)), \(c\) is the speed of light (\(2.998 \times 10^{8} \mathrm{m/s}\)), and \(\lambda\) is the wavelength of the light (\(5.6 \times 10^{-7} \mathrm{m}\)). Using these values, we can find the energy of a single photon: \(E = \dfrac{(6.626 \times 10^{-34} \mathrm{Js})(2.998 \times 10^{8} \mathrm{m/s})}{5.6 \times 10^{-7} \mathrm{m}} \)

To find the total energy entering the eye, we will multiply the intensity of the light by the area of the aperture of the human eye (which can be approximated as a circle): \(\mathrm{Total~Energy} = \mathrm{Intensity} \times \mathrm{Area} \times \mathrm{Time}\) The area of the aperture can be calculated using the formula \(A = \pi r^2\), where \(A\) is the area and \(r\) is the radius of the eye (given as \(2.5 \times 10^{-3} \mathrm{m}\)). First, calculate the area of the aperture: \(A = \pi (2.5 \times 10^{-3} \mathrm{m})^2\) Now, calculate the total energy entering the eye in 1 second: \(\mathrm{Total~Energy} = (1.6 \times 10^{-9} \mathrm{Wm}^{-2}) (\pi (2.5 \times 10^{-3} \mathrm{m})^2)(1 \mathrm{s})\)

Lastly, we will find the number of photons entering the eye per second by dividing the total energy by the energy of a single photon: \(\mathrm{Number~of~Photons\, =\, \dfrac{Total\,~Energy}{Energy\,~of\,~a\,~Single\,~Photon}}\) \(\mathrm{Number~of ~Photons~ in ~1s} = \dfrac{(1.6 \times 10^{-9} \mathrm{Wm}^{-2}) (\pi (2.5 \times 10^{-3} \mathrm{m})^2)(1 \mathrm{s})}{\dfrac{(6.626 \times 10^{-34} \mathrm{Js})(2.998 \times 10^{8} \mathrm{m/s})}{5.6 \times 10^{-7} \mathrm{m}}}\) After calculating the number of photons, we find the closest answer is: (A) \(7.8 \times 10^{4} \mathrm{~s}^{-1}\)