A 60 ,W source emits monochromatic light of w | vedclass

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

Question

(B) Given: Power of the source $P = 60 \,W$.Wavelength of light $\lambda = 662.5 \,nm = 6.625 	imes 10^{-7} \,m$.The energy of a single photon is giv

Vedclass · Accepted Answer

$2 	imes 10^{20}$

Vedclass · Answer

(B) Given: Power of the source $P = 60 \,W$.Wavelength of light $\lambda = 662.5 \,nm = 6.625 	imes 10^{-7} \,m$.The energy of a single photon is given by $E = \frac{hc}{\lambda}$.Substituting the values ($h = 6.625 	imes 10^{-34} \,J \cdot s$ and $c = 3 	imes 10^8 \,m/s$):$E = \frac{6.625 	imes 10^{-34} 	imes 3 	imes 10^8}{6.625 	imes 10^{-7}} = 3 	imes 10^{-19} \,J$.The number of photons emitted per second $(n)$ is given by the ratio of total power to the energy of one photon:$n = \frac{P}{E} = \frac{60}{3 	imes 10^{-19}} = 20 	imes 10^{19} = 2 	imes 10^{20}$ photons/second.

$A$ $60 \,W$ source emits monochromatic light of wavelength $662.5 \,nm$. The number of photons emitted per second is

Similar Questions

$A$ photon creates an electron-positron pair with equal kinetic energy. If the kinetic energy of each particle is $0.29 \ MeV$,the energy of the photon must be .......... $MeV$.

The value of Planck's constant is

The ratio of the energy of a photon with $\lambda = 150\,nm$ to that with $\lambda = 300\,nm$ is

Write the equation for the effective mass of a photon.

$A$ gamma ray photon creates an electron-positron pair. If the rest mass energy of an electron is $0.5\, MeV$ and the total kinetic energy $(K.E.)$ of the electron-positron pair is $0.78\, MeV$,then the energy of the gamma ray photon must be.......$MeV$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module