| Given E (in eV) \[=\frac{1237}{\lambda (in\,nm)}\] |
A) 1.5 eV
B) 4.5 eV
C) 15.1 eV
D) 3.0 eV
Correct Answer: A
Solution :
\[{{K}_{\max }}=\frac{hc}{\lambda }-\frac{hc}{{{\lambda }_{0}}}\] \[\Rightarrow {{K}_{\max }}=hc\left( \frac{{{\lambda }_{0}}-\lambda }{\lambda {{\lambda }_{0}}} \right)\] \[\Rightarrow {{K}_{\max }}=(1237)\left( \frac{380-260}{380\times 260} \right)\]\[=1.5eV\]
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