A) \[0.33\times {{10}^{6}}\]
B) \[7\times {{10}^{-24}}\]
C) \[{{10}^{-22}}\]
D) \[5\times {{10}^{-22}}\]
Correct Answer: D
Solution :
Energy of photon is given by \[E=\frac{hc}{\lambda }\] ??(i) where h is Plancks constant, c the velocity of light and \[\lambda \] its wavelength. de-Broglie wavelength is given by \[\lambda =\frac{h}{p}\] ??.(ii) p being momentum of photon. From Eqs. (i) and (ii), we can have \[E=\frac{hc}{h/p}=pc\] or \[p=E/c\] Given, \[E=1MeV=1\times {{10}^{6}}\times 1.6\times {{10}^{-19}}J\] \[c=3\times {{10}^{8}}m/s\] Hence, after putting numerical values, we obtain \[p=\frac{1\times {{10}^{6}}\times 1.6\times {{10}^{-19}}}{3\times {{10}^{8}}}kg-m/s\] \[=5\times {{10}^{-22}}kg-m/s\]
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