A) 0.58 eV
B) 2.48 eV
C) 1.48 eV
D) 2.68 eV
Correct Answer: A
Solution :
Energy of photon is given by |
\[E=\frac{hc}{\lambda }=\frac{12375}{\lambda \,({\AA})}eV\] |
\[\therefore \] \[E=\frac{12375}{5000}=2.48\,eV\] |
Einsteins photoelectric equation is |
\[{{E}_{k}}=E-W\] |
\[=2.48\,eV-1.9\,eV\] |
\[=0.58\,eV\] |
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