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