A) 15keV
B) 1.5 keV
C) 150 keV
D) 1.5MeV
Correct Answer: A
Solution :
The de-Broglie wavelength \[(\lambda )\] is given by \[\lambda =\frac{h}{p}=\frac{h}{mv}\] where h is Plancks constant, m the mass and v the velocity. Given, \[\lambda =10\,pm={{10}^{-11}}m\], \[m=9.1\times {{10}^{-31}}kg\], \[h=6.6\times {{10}^{-34}}J-s\]. \[\therefore \] \[v=\frac{h}{m\,\lambda }=\frac{6.6\times {{10}^{-34}}}{9.1\times {{10}^{-31}}\times {{10}^{-11}}}\] \[\Rightarrow \] \[v=7.25\times {{10}^{7}}m/s\] Also, kinetic energy is the energy possessed due to velocity (v) is given by \[KE=\frac{1}{2}m\,{{v}^{2}}\] Given, \[m=9.1\times {{10}^{-31}}kg\], \[v=7.25\times {{10}^{7}}\,m/s\] \[\therefore \] \[KE=\frac{1}{2}\times 9.1\times {{10}^{-31}}\times {{(7.25\times {{10}^{7}})}^{2}}\] Since, \[1\,eV=1.6\times {{10}^{-19}}J\] \[\therefore \] \[KE=\frac{1}{2}\times 9.1\times {{10}^{-31}}\times \frac{{{(7.25\times {{10}^{7}})}^{2}}}{1.6\times {{10}^{-19}}}\] = 15 keV
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