A) \[6.62\times {{10}^{-48}}\]
B) \[6.62\times {{10}^{-20}}\]
C) \[\frac{6.62}{3}\times {{10}^{-28}}\]
D) \[3\times 6.62\times {{10}^{-28}}\]
Correct Answer: B
Solution :
\[E=h\nu =6.64\times {{10}^{-34}}\times 1.0\times {{10}^{14}}\]\[=6.62\times {{10}^{-20}}J\]
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