A 160 watt light source is radiating light of wavelength 6200 Å uniformly in all directions. The photon flux at a distance of 1.8 m is of the order of (Planck's constant `6.63xx 10^(-34) J-s)`

Options

A. `10^(2)m^(-2)s^(-1)`
B. `10^(12)m^(-2)s^(-1)`
C. `10^(19)m^(-2)s^(-1)`
D. `10^(25)m^(-2)s^(-1)`

Answer

C

Explanation

Intensity of light at `1.8m=(P)/(4pi(1.8)^(2))`
implies `I=(160)/(4xxpixx(1.8)^(2))`
Photon flux = No. of photon per unit area
= `(I)/(hc//lambda)=(Ilambda)/(hc)=(160xx6200xx10^(-10))/(4xxpixx(1.8)^(2)xx6.63xx10^(-34)xx3xx10^(8))`
= `1.22xx10^(19)~~10^(19)`