A) 4.58 eV
B) 2.29 eV
C) 1.14 eV
D) 0.57 eV
Correct Answer: B
Solution :
\[{{W}_{0}}=\frac{12375}{{{\lambda }_{0}}({\AA})}=\frac{12375}{5420}=2.28\,eV\]
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