A source S_1 is producing, 10^{15} photons pe | vedclass

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

Question

For a source $S_{1}$

Wavelength, $\lambda_{1}=5000 \,\,{\mathop {	ext{A}}\limits^o }$

Number of photons emitted per second, $N_{1}=10^{15}$

Vedclass · Accepted Answer

$1$

Vedclass · Answer

For a source $S_{1}$

Wavelength, $\lambda_{1}=5000 \,\,{\mathop {	ext{A}}\limits^o }$

Number of photons emitted per second, $N_{1}=10^{15}$

Energy of each photon, $E_{1}=\frac{h c}{\lambda_{1}}$

Power of source $S_{1}, P_{1}=E_{1} N_{1}=\frac{N_{1} h c}{\lambda_{1}}$

For a source $S_{2}$

Wavelength, $\lambda_{2}=5100\,\,{\mathop {	ext{A}}\limits^o }$

Number of photons emitted per second,

$N_{2}=1.02 	imes 10^{15}$

Energy of each photon, $E_{2}=\frac{h c}{\lambda_{2}}$

Power of source $S_{2},\,\,P_{2}=N_{2} E_{2}=\frac{N_{2} h c}{\lambda_{2}}$

$	herefore \,\,\frac{{{	ext{ Power of}}\,{	ext{ }}{S_2}}}{{{	ext{ Power of }}\,{S_1}}} = \frac{{{P_2}}}{{{P_1}}}$ $ = \frac{{\frac{{{N_2}hc}}{{{\lambda _2}}}}}{{\frac{{{N_1}hc}}{{{\lambda _1}}}}} = \frac{{{N_2}{\lambda _1}}}{{{N_1}{\lambda _2}}}$

$ = \frac{{(1.02 	imes {{10}^{15}}{	ext{ photons/s) }} 	imes (5000{\mkern 1mu} {\mkern 1mu} \mathop {	ext{A}}\limits^o )}}{{({{10}^{15}}{	ext{ photons/s) }} 	imes (5100{\mkern 1mu} {\mkern 1mu} \mathop {	ext{A}}\limits^o )}}$ $ = \frac{{51}}{{51}} = 1$

A source $S_1$ is producing, $10^{15}$ photons per second of wavelength $5000 \;\mathring A.$ Another source $S_2$ is producing $1.02 \times 10^{15}$ photons per second of wavelength $5100\;\mathring A$. Then, $($ power of $S_2)/$ $($ power of $S_1)$ is equal to

Similar Questions

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

A photon of $1.7 \times {10^{ - 13}}$ Joules is absorbed by a material under special circumstances. The correct statement is

The threshold wavelength for photoelectric emission from a material is $5200\, \mathring A$. Photo-electrons will be emitted when this material is illuminated with monochromatic radiation from a

Photo-electric effect can be explained by

An AIR station is broadcasting the waves of wavelength $300$ metres. If the radiating power of the transmitter is $10 kW$, then the number of photons radiated per second is