A) \[K+{{E}_{0}}\]
B) 2 K
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., \[hv=\frac{1}{2}m{{v}^{2}}+{{E}_{0}}\] |
| or \[hv={{K}_{\max }}+{{E}_{0}}\] (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
