A) \[K+{{E}_{0}}\]
B) \[2K\]
C) \[K\]
D) \[K+hv\]
Correct Answer: D
Solution :
Key Idea: The energy of photon is used in liberating the electron from metal surface and in imparting the kinetic energy to emitted photoelectron. According to Einstein's photoelectric effect energy of photon = KE of photoelectron + work function of metal \[i.e.,hv=\frac{1}{2}m{{v}^{2}}+{{E}_{0}}\] \[orhv={{K}_{\max }}+{{E}_{0}}...(i)\] Now, we have given, \[v'=2v\] Therefore, \[K{{'}_{\max }}=h\,(2v)-{{E}_{0}}\] \[K{{'}_{\max }}=2hv-{{E}_{0}}....(ii)\] From Eqs. (i) and (ii), we have \[K{{'}_{\max }}=2\,({{K}_{\max }}+{{E}_{0}})-{{E}_{0}}\] \[=2\,{{K}_{\max }}+{{E}_{0}}\] \[={{K}_{\max }}+({{K}_{\max }}+{{E}_{0}})\] \[={{K}_{\max }}+hv[From\,\,Eq.\,(i)]\] putting \[{{K}_{\max }}=K\] \[\therefore K{{'}_{\max }}=K+hv\] Note: The photoelectric emission is an instantaneous process. The time lag between the incidence of radiations and emission of photoelectrons is very small, less than even \[{{10}^{-9}}\] second.
You need to login to perform this action.
You will be redirected in
3 sec
