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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