A) \[\sqrt{2}\,{{v}_{0}}\]
B) \[2\,{{v}_{0}}\]
C) \[2\sqrt{2}\,{{v}_{0}}\]
D) \[4\,{{v}_{0}}\]
Correct Answer: B
Solution :
\[{{v}_{\max }}=\sqrt{\frac{2h\,(v-{{v}_{0}})}{m}}\] \[v=2{{v}_{0}}\] \[\therefore \] \[{{v}_{\max }}={{v}_{0}}=\sqrt{\frac{2h\,(2{{v}_{0}}-{{v}_{0}})}{m}}\] \[{{v}_{0}}=\sqrt{\frac{2h{{v}_{0}}}{m}}\] Again \[v=5{{v}_{0}}\] \[\therefore \] The maximum velocity of the photoelectron \[{{v}_{\max }}=\sqrt{\frac{2h(5{{v}_{0}}-{{v}_{0}})}{m}}\] \[=\sqrt{\frac{2h4{{v}_{0}}}{m}}\] \[=2\sqrt{\frac{2h{{v}_{0}}}{m}}=2\,{{v}_{0}}\]
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