A) 3 eV
B) 4.5 eV
C) 6 eV
D) 9 eV
Correct Answer: C
Solution :
\[{{(KE)}_{\max }}=hv-\text{o}{{|}_{0}}\] So, \[1\,eV=h{{v}_{0}}-\text{o}{{|}_{0}}\] ?(i) and \[3\,eV=\frac{h{{v}_{0}}}{2}-\text{o}{{|}_{0}}\] ?(ii) \[\Rightarrow \] \[3eV-1eV=\frac{h{{v}_{0}}}{2}\] or \[h{{v}_{0}}=4\,eV\] From Eq.(i), \[\text{o}{{|}_{0}}=h{{v}_{0}}-1\,eV\] \[=4eV-1\,eV\] \[=3\,eV\] \[\therefore \] \[{{(KE)}_{\max }}=h\times \frac{9{{v}_{0}}}{4}-3\,eV\] \[=\frac{9}{4}(4\,eV)-3\,eV\] \[=6\,eV\]You need to login to perform this action.
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