A photodiode sensor is used to measure the output of a 300 W lamp kept 10 m away. The sensor has an opening of 2 cm in diameter.
How many photons enter the sensor if the wavelength of the light is 660 nm and the exposure time is 100 ms. (Assume that all the energy of the lamp is given off as light and `h = 6.6 xx 10^(-34) Js`)

Options

A. `3.6 xx 10^(13) `
B. `2.8 xx 10^(13)`
C. `2.5 xx 10^(13)`
D. `1.8 xx 10^(13)`

Answer

C

Explanation

Given that, power of lamp, P = 300 W distance of lamp from sensor d = 10 m
radius of sensor opening , r=1 cm , wavelenght of light emitted from photodiode, `lambda = 660 nm` and exposure time `t= 100 ms = 100 xx 10^(-3)s`
Engergy of photons, `E= (hc)/(lambda)`
Putting the given values we get
`rArr E = (6.6 xx10^(-34)xx3xx10^8)/(660 nm) = 3xx1-^(-19)`
The area of exposure of lamp at a radius of 10 m
`A_0 = 4pi r^2 = 4pi xx 10^2 m^2`
Similarly the are of sensor
`A^'s = pir^2 = pixx(10^(-2))^2 = pi xx 10^(-4) m^2`
Energy emitted by lamp in exposure time
`E_0` = power of lamp (p) `xx` exposure time(t)
`E_0 = 300 xx 100 xx10^(-3) = 3J`
So, the number of photons entering to sensor is given by
`N = (E_0) /E xx(A_s^') /(A_0)`
`= (30)/(3xx10^(-19))xx(pixx10^(-4))/(4pixx10^2)`
`rArr N = 2.5 xx 10^(13)`