A) 0.58 eV
B) 2.48 eV
C) 1.24 eV
D) 1.18 eV
Correct Answer: A
Solution :
The energy of photon\[E=\frac{hc}{\lambda }\] \[=\frac{12375}{\lambda }eV\] \[E=\frac{12375}{5000}\] \[=2.48eV\] The kinetic energy of equation of photoelectric \[{{E}_{k}}=E-W\] \[=0.58\,eV\]
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