$\begin {array} {1 1} (a)\;1.008 \times 10^4s^{-1} & \quad (b)\;10.08 \times 10^3s^{-1} \\ (c)\;1.008 \times 10^3s^{-1} & \quad (d)\;10.08 \times 10^4s^{-1} \end {array}$

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

Ans : (a)

$I=10^{-10} \large\frac{W}{m^2}$

$A = 0.4 cm^2 = 0.4 \times 10^{-4} m^2 \: \: \: \: v=6 \times 10^{14}Hz$

So, $E=hv = 6.6 \times 10^{-34} \times 6 \times 10^{14} Js$

$= 3.96 \times 10^{-19} J$

Let n be total number of photons falling per second, per unit area of the pupil.

So, total energy per unit for n falling photons : $E = n \times 3.96 \times 10^{-19} J/sm^2$

Energy per unit area per second is intensity of light.

$E=I$

$ \Rightarrow n \times 3.96 \times 10^{-19} = 10^{-10}$

$ \Rightarrow n= \large\frac{10^{-10}}{3.96 \times 10^{-19} } = 2.52 \times 10^8 m^2 s^{-1}$

So, total no. of photons entering per second is : $n_A = n \times A$

$= 2.52 \times 10^8 \times 0.4 \times 10^{-4}$

$= 1.008 \times 10^4 s^{-1}$

Ask Question

Take Test

x

JEE MAIN, CBSE, AIPMT Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...