A 160 ,W light source is radiating light of w | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(c)

Photon flux $=$ Number of photons per unit area per unit time

$=\frac{I}{\left(\frac{h c}{\lambda}\right)}=\frac{I \lambda}{h c}=\frac{P \la

Vedclass · Accepted Answer

$10^{19}$

Vedclass · Answer

(c)

Photon flux $=$ Number of photons per unit area per unit time

$=\frac{I}{\left(\frac{h c}{\lambda}ight)}=\frac{I \lambda}{h c}=\frac{P \lambda}{4 \pi r^2 h c}$

$=\frac{160 	imes 6200 	imes 10^{-10}}{4 	imes \pi 	imes(1.8)^2 	imes 6.63 	imes 10^{-34} 	imes 3 	imes 10^8}$

$=1.22 	imes 10^{19} \,m ^{-2} s ^{-1}$

A $160 \,W$ light source is radiating light of wavelength $6200 \,\mathring A$ uniformly in all directions. The photon flux at a distance of $1.8 \,m$ is of the order of .......... $m ^{-2} s ^{-1}$ (Planck's constant $\left.=6.63 \times 10^{-34} \,J - s \right)$

Similar Questions

Dual nature of radiation is shown by

The momentum of a photon with energy $20\, eV$ is

A $200\, W$ sodium street lamp emits yellow light of wavelength $0.6\, \mu \,m$. Assuming it to be $50\%$ efficient in converting electrical energy to light, the number of photon of yellow light it emits per second is

Photo-electric effect can be explained by

When light falls on a metal surface, the maximum kinetic energy of the emitted photo-electrons depends upon