A) 1.5 keV
B) 15 keV
C) 150 keV
D) 1.5 MeV
Correct Answer: B
Solution :
From de-Broglie equation, we have \[\lambda =\frac{h}{p}=\frac{h}{m\nu }\] where \[\lambda \] is the wavelength, h is the Planck?s constant, m is the mass, p is momentum and v is the velocity. Given, \[\lambda = 10 pm = 1{{0}^{-11}}m,\,m =9.1 \times 1{{0}^{-}}^{31}\,kg,\] \[h=6.6\times {{10}^{-34}}Js\] \[\therefore \,\,\,\,\,\,\,\nu =\frac{h}{m\lambda }\] \[=\,\,\,\,\,\frac{6.6\times {{10}^{-34}}}{9.1\times {{10}^{-31}}\times {{10}^{-11}}}\] \[=\,\,\,\,7.25\,\,\times \,\,1{{0}^{7}}\,m/s\] Energy of electron \[=\,\,\,\frac{1}{2}m{{\nu }^{2}}=\frac{1}{2}\times \frac{9.1\times {{10}^{-31}}\times {{(7.25\times {{10}^{7}})}^{2}}}{1.6\times {{10}^{-19}}}\] \[=\,\,15ke\,V\]You need to login to perform this action.
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