A) \[2\,eV\]
B) \[4\,eV\]
C) \[14\,eV\]
D) \[18\,eV\]
Correct Answer: C
Solution :
[c] \[K{{E}_{{{\max }_{1}}}}=\frac{hC}{{{\lambda }_{1}}}-\phi \] Now \[\phi =\frac{KE{{\max }_{1}}}{3}=2eV\] (given) So \[\frac{hC}{{{\lambda }_{1}}}=6+2=8eV\] and \[{{\lambda }_{1}}{{T}_{1}}={{\lambda }_{2}}{{T}_{2}}\] So \[\frac{hC}{{{\lambda }_{2}}}=2\frac{hC}{{{\lambda }_{1}}}=16eV\] \[K{{E}_{{{\max }_{2}}}}=\frac{hC}{{{\lambda }_{2}}}-\phi \] \[K{{E}_{\max 2}}=16-2=14eV\]You need to login to perform this action.
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