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 Planck?s constant, c the velocity of light and 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 = 1 MeV = \[1\times {{10}^{6}}\times 1.6\times {{10}^{-19}}\,J\] \[c=3\times {{10}^{8}}\,m/s\,\] Hence, after putting numerical values, we obtain \[p=\frac{1\times {{10}^{6}}\times 1.6\times {{10}^{-19}}}{3\times {{10}^{8}}}\,kgm/s\] \[=5\times {{10}^{-22}}\,kgm/s\]
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