A) 17\[eV\]
B) 22\[eV\]
C) 27\[eV\]
D) 37\[eV\]
Correct Answer: C
Solution :
\[E=h\nu =6.6\times {{10}^{-34}}\times 8\times {{10}^{15}}=5.28\times {{10}^{-18}}J=33eV\] By using \[E={{W}_{0}}+{{K}_{\max }}\Rightarrow {{K}_{\max }}=E-{{W}_{0}}\] \[=33-6.125=27eV\]
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