A) 3.72 eV
B) 6.72 eV
C) 5.72 eV
D) 4.67eV
Correct Answer: D
Solution :
From Einsteins theory, maximum kinetic energy \[({{E}_{k}})\]of emitted photoelectron is given by \[{{E}_{k}}=hv-W\] Where W is work Function of silver,\[v\]is frequency. \[\Rightarrow \] \[W=\frac{hc}{\lambda }-{{E}_{k}}\] Putting \[h=6.6\times {{10}^{-34}}\,J-s,\] \[c=3\times {{10}^{8}}\,m/s,\,\lambda ={{10}^{-7}}\,m\] \[{{E}_{k}}=7.7\,eV\] \[=7.7\times 1.6\times {{10}^{-19}}V\] \[\therefore \] \[W=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{{{10}^{-7}}}\] \[-7.7\times 1.6\times {{10}^{-19}}J\] \[\Rightarrow \]\[W=19.8\times {{10}^{-19}}-12.32\times {{10}^{-19}}\] \[\Rightarrow \] \[W=7.48\times {{10}^{-19}}\] \[W=\frac{7.48\times {{10}^{-19}}}{1.6\times {{10}^{-19}}}eV\] \[=4.67\,eV\]
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