The momentum of a photon of energy `1 MeV` in `kg m//s` will be

Options

A. `0.33xx10^(6)`
B. `7xx10^(-24)`
C. `10^(-22)`
D. `5xx10^(-22)`

Answer

D

Explanation

Energy of photon is given by
`E=(hc)/(lamda)` . . . (i)
where, h is the planck's constant, c is the velocity of light and `lamda` its wavelength
de-Broglie wavelength is given by
`lamda=(h)/(p)` . . .(ii)
Where, p is being momentum of photon.
From eqs. (i) and (ii), we get
`E=(hc)/(h//p)=pc implies p=E//c`
Gvien, `E=1MeV=1xx10^(6)xx1.6xx10^(-19)J`,
`c=3xx10^(8)ms^(-1)`
Hence, after putting numberical values, we obtain
`p=(1xx10^(6)xx1.6xx10^(-19))/(3xx10^(8))=5xx10^(-22)kg-ms^(-1)`.