A) \[\frac{hc}{2\lambda }\]
B) \[\frac{hc}{\lambda }\]
C) \[\frac{2hc}{\lambda }\]
D) \[\frac{hc}{3\lambda }\]
Correct Answer: A
Solution :
| According to Einstein's photoelectric equation, |
| \[E={{K}_{\max }}+\phi \] |
| where, \[{{K}_{\max }}\] is maximum kinetic energy of emitted electron and \[\phi \] is work function of an electron. |
| \[{{K}_{\max }}=E-\phi =hv-\phi \] |
| \[{{K}_{\max }}=\frac{hc}{\lambda }-\phi \] (i) |
| Similarly, in second case, maximum kinetic energy of emitted electron is 3 times that in first case, we get |
| \[3{{K}_{\max }}=\frac{hc}{\frac{\lambda }{2}}-\phi \] (ii) |
| Solving Eqs. (i) and (ii), we get work function of an emitted electron from a metal surface. |
| \[\phi =\frac{hc}{2\lambda }\] |
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