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

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(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

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$5 	imes 10^{-22}$

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(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

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