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The momentum of a photon of energy $1MeV$ in kgm/s, will be

$\begin {array} {1 1} (a)\;0.33 \times 10^6 & \quad (b)\;7\: v\: 10^{-24} \\ (c)\;10^{-22} & \quad (d)\;5 \times 10^{-22} \end {array}$

 

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And : (d) $5 \times 10^{-22}$
Energy of a photon, $E = hc/\lambda = hc/(h/p) \: \: \: \: (\lambda = h/p)$
$= pc$
$ \Rightarrow p = E/c$
$ \Rightarrow p = 10^6 \times 1.6 \times 10^{-19} / (3 \times 10^8 ) kgm/s$
$ \Rightarrow p = 5 \times 10^{-22} kgm/s$

 

answered Dec 23, 2013 by thanvigandhi_1
edited Mar 13, 2014 by thanvigandhi_1
 

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