A) 1.5 KeV
B) 15 KeV
C) 150 KeV
D) 1.5 KeV
Correct Answer: B
Solution :
Wave length of the electron wave be \[10\times {{10}^{-12}}m\], Using \[\lambda =\frac{h}{\sqrt{2mE}}\]\[\Rightarrow E=\frac{{{h}^{2}}}{{{\lambda }^{2}}\times 2m}\] \[=\frac{{{(6.63\times {{10}^{-34}})}^{2}}}{{{(10\times {{10}^{-12}})}^{2}}\times 2\times 9.1\times {{10}^{-31}}}Joule\] \[=\frac{{{(6.63\times {{10}^{-34}})}^{2}}}{{{(10\times {{10}^{-12}})}^{2}}\times 2\times 9.1\times {{10}^{-31}}\times 1.6\times {{10}^{-19}}}eV\] = 15.1 KeV.
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