A) \[2\times {{10}^{-15}}J\]
B) \[2\times {{10}^{-16}}J\]
C) \[2\times {{10}^{-17}}J\]
D) \[2\times {{10}^{-18}}J\]
Correct Answer: A
Solution :
[a] \[E=hc/\lambda =6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}/1\times {{10}^{-10}}\] \[2\times {{10}^{-15}}J\]
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