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, \[E=1\text{ }MeV=1\times {{10}^{6}}\times 1.6\times {{10}^{-19}}\,J\] |
| \[c=3\times {{10}^{8}}\,m/s\,\] |
| 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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