A) \[K+{{E}_{0}}\]
B) 2K
C) K
D) \[K+hv\]
Correct Answer: D
Solution :
According to Einstein's photoelectric effect energy of photon = KE of photoelectron + word function of metal That is, \[hv=\frac{1}{2}m{{v}^{2}}+{{E}_{0}}\] \[hv=\frac{1}{2}m{{v}^{2}}+{{E}_{0}}\] or \[hv={{K}_{\max }}+{{E}_{0}}\] Now, we have given, \[v'=2v\] Therefore, \[K_{\max }^{'}=2hv-{{E}_{0}}\] 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 \[{{K}_{\max }}=K\Rightarrow K_{\max }^{'}=K+hv\]
You need to login to perform this action.
You will be redirected in
3 sec
