The momentum of a photon of energy 1 text{ Me | vedclass

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

Question

(D) The energy of a photon is given by $E = \frac{hc}{\lambda}$,where $h$ is Planck's constant,$c$ is the speed of light,and $\lambda$ is the waveleng

Vedclass · Accepted Answer

$5 	imes 10^{-22}$

Vedclass · Answer

(D) The energy of a photon is given by $E = \frac{hc}{\lambda}$,where $h$ is Planck's constant,$c$ is the speed of light,and $\lambda$ is the wavelength.According to the de-Broglie relation,the momentum $p$ is given by $p = \frac{h}{\lambda}$,which implies $\lambda = \frac{h}{p}$.Substituting $\lambda$ into the energy equation: $E = \frac{hc}{h/p} = pc$.Therefore,the momentum is $p = \frac{E}{c}$.Given: $E = 1 	ext{ MeV} = 1 	imes 10^{6} 	imes 1.6 	imes 10^{-19} 	ext{ J} = 1.6 	imes 10^{-13} 	ext{ J}$ and $c = 3 	imes 10^{8} 	ext{ m/s}$.Substituting these values: $p = \frac{1.6 	imes 10^{-13}}{3 	imes 10^{8}} \approx 0.533 	imes 10^{-21} 	ext{ kg-m/s} = 5.33 	imes 10^{-22} 	ext{ kg-m/s}$.Rounding to the nearest provided option,we get $p = 5 	imes 10^{-22} 	ext{ kg-m/s}$.

The momentum of a photon of energy $1 \text{ MeV}$ in $\text{kg-m/s}$ will be

Similar Questions

The fact that photons carry energy was established by

$A$ photon of frequency $v$ has a momentum associated with it. If the speed of light is $c$,then the momentum is .......

When a photon enters glass from air,which one of the following quantities does not change?

$A$ 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,$(\text{power of } S_2) / (\text{power of } S_1)$ is equal to

Which of the following has a mass closest in value to that of the positron? $(1\;a.m.u = 931\;MeV)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module