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. \[\phi =\frac{hc}{2\lambda }\]
You need to login to perform this action.
You will be redirected in
3 sec
