A) \[3.72\,\,eV\]
B) \[6.72\,\,eV\]
C) \[5.72\,\,eV\]
D) \[4.67\,\,eV\]
Correct Answer: D
Solution :
From Einstein's 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\[h=6.6\times {{10}^{-34}}J\text{-}s\] \[c=3\times {{10}^{8}}m/s,\,\,\lambda ={{10}^{-7}}m\] \[{{E}_{k}}=7.7\,\,eV\] \[=7.7\times 1.6\times {{10}^{-19}}V\] \[\therefore \] \[W=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{{{10}^{-7}}}\] \[-7.7\times 1.6\times {{10}^{-19}}J\] \[\Rightarrow \] \[W=19.8\times {{10}^{-19}}-12.32\times {{10}^{-19}}\] \[\Rightarrow \] \[W=7.48\times {{10}^{-19}}\] \[W=\frac{7.48\times {{10}^{-19}}}{1.6\times {{10}^{-19}}}eV\] \[=4.67\,\,eV\]
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