A) 1.5 keV
B) 15 keV
C) 150 keV
D) 1.5 MeV
Correct Answer: B
Solution :
From de-Broglie equation, we have \[\lambda =\frac{h}{p}=\frac{h}{mv}\] where\[\lambda \]is wavelength, h is Plancks constant, m is mass, p is momentum and \[v\]is velocity. Given, \[\lambda =10\,pm={{10}^{-11}}\,m,\] \[m=9.1\times {{10}^{-31}}kg,\,h=6.6\times {{10}^{-34}}\,J-s\] \[\therefore \] \[v=\frac{h}{m\lambda }\] \[=\frac{6.6\times {{10}^{-34}}}{9.1\times {{10}^{-31}}\times {{10}^{-11}}}\] \[=7.25\times {{10}^{7}}\,m/s\] Energy of electron \[=\frac{1}{2}m{{v}^{2}}=\frac{1}{2}\times \frac{9.1\times {{10}^{-31}}\times {{(7.25\times {{10}^{7}})}^{2}}}{1.6\times {{10}^{-19}}}\] \[=15\,keV\]
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