A) \[{{E}_{\max }}=hv-\phi \]
B) \[E=m{{c}^{2}}\]
C) \[{{E}^{2}}={{p}^{2}}{{c}^{2}}+m_{0}^{2}{{c}^{4}}\]
D) \[E=\left( \frac{1}{2} \right)m{{v}^{2}}\]
Correct Answer: A
Solution :
Einstein's photo electronic equation is \[{{E}_{\max }}=hv-\phi \]
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