A) \[K+{{e}_{0}}\]
B) \[2K\]
C) \[K\]
D) \[K+hv\]
Correct Answer: D
Solution :
According to Einsteins photoelectric effect energy of photon = KE of photoelectron \[+\] work function of metal ie, \[hv=\frac{1}{2}m{{v}^{2}}+{{E}_{0}}\] or \[hv={{K}_{\max }}+{{E}_{0}}\] ?.(i) Now, we have given, \[v=2v\] Therefore, \[K{{}_{\max }}=2hv-{{E}_{0}}\] From Eqs. (i) and (ii) we have 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 \[{{K}_{\max }}=K\] \[\therefore \] \[K{{}_{\max }}=K+hv\]
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