A) \[11.2\times {{10}^{-18}}J\]
B) \[12.9\times {{10}^{-16}}J\]
C) \[1.7\times {{10}^{-15}}J\]
D) \[10\times {{10}^{-16}}J\]
Correct Answer: B
Solution :
Required ionisation energy\[E=\frac{hc}{\lambda }\] \[=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{1.54\times {{10}^{-10}}}\] \[=12.9\times {{10}^{-16}}J\]
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