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. |
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