A) \[3\,h{{\nu }_{0}}\]
B) \[2\,h{{\nu }_{0}}\]
C) \[\frac{3}{2}\,h{{\nu }_{0}}\]
D) \[\frac{1}{2}h{{\nu }_{0}}\]
Correct Answer: A
Solution :
\[E=h{{\nu }_{0}}+{{K}_{\max }}\Rightarrow h(4{{\nu }_{0}})=h{{\nu }_{0}}+{{K}_{\max }}\]\[\Rightarrow {{K}_{\max }}=3\,h{{\nu }_{0}}\].
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