A) 0.168 eV
B) 16.8 eV
C) 1.68 eV
D) 2.5 eV
Correct Answer: B
Solution :
[b] \[\lambda =\frac{h}{\sqrt{2mE}}\Rightarrow E=\frac{{{h}^{2}}}{2m{{\lambda }^{2}}}\] \[=\frac{{{(6.6\times {{10}^{-34}})}^{2}}}{2\times 9.1\times {{10}^{-31}}\times {{(0.3\times {{10}^{-9}})}^{2}}}=2.65\times {{10}^{-18}}J\] \[=16.8\,eV\]
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