A) 0.58 eV
B) 2.48 eV
C) 1.24 eV
D) 1.16 eV
Correct Answer: A
Solution :
Energy of photon is given by \[E=\frac{hc}{\lambda }=\frac{12375}{\lambda \,({\AA})}eV\] \[\therefore E=\frac{12375}{5000}=2.48\,eV\] Einstein?s photoelectric equation is \[{{E}_{k}}=E-W\] \[=2.48\,eV-1.9\,eV\] \[=0.58\,eV\]You need to login to perform this action.
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