A) 2.5V
B) 5V
C) 2.5 kV
D) 5kV
Correct Answer: B
Solution :
\[\lambda =\frac{h}{\sqrt{2eVm}}\] \[5\overset{\text{o}}{\mathop{\text{A}}}\,=\frac{6.6\times {{10}^{-34}}}{\sqrt{2\times 1.6\times {{10}^{-19}}\times 9.1\times {{10}^{-31}}\times V}}\] \[5\overset{\text{o}}{\mathop{\text{A}}}\,=\frac{6.6\times {{10}^{-34}}}{5.4\times {{10}^{-25}}\sqrt{V}}\] \[V={{\left( \frac{6.6\times {{10}^{-34}}}{5.4\times {{10}^{-25}}\times 5\times {{10}^{-10}}} \right)}^{2}}\] \[V=5.76volt\]
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