A) 540nm
B) 400 nm
C) 310nm
D) 220 nm
Correct Answer: C
Solution :
\[W=h{{v}_{0}}=\frac{hc}{{{\lambda }_{0}}}\] Wave function in electron volt\[{{W}_{0}}(eV)\] \[{{W}_{0}}=\frac{12400}{{{\lambda }_{0}}(\overset{o}{\mathop{\text{A}}}\,)}eV\] \[4.0=\frac{12400}{{{\lambda }_{0}}}\] \[{{\lambda }_{0}}=310\,nm\]
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