A) \[0.33 \times 1{{0}^{6}}\]
B) \[7 \times 1{{0}^{-24}}\]
C) \[1{{0}^{-22}}\]
D) \[5 \times 1{{0}^{-22}}\]
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 \[\lambda \] 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, \[E = 1 Me\,V = 1 \times 1{{0}^{6}}\times 1.6 \times 1{{0}^{-}}^{19}J\] \[c=3\times {{10}^{8}}\,m/s\] 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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