A) 2.0 V
B) 0.5 V
C) 1.0 V
D) 1.5 V
Correct Answer: C
Solution :
\[{{E}_{1}}=\frac{1240eV-nm}{300}=4.13eV\] \[{{E}_{2}}=\frac{1240\,eV-nm}{400}=3.10\,eV\] \[{{K}_{{{\max }_{1}}}}={{E}_{1}}-{{\phi }_{0}},{{K}_{{{\max }_{2}}}}={{E}_{2}}-{{\phi }_{0}}\] \[{{K}_{{{\max }_{1}}}}-{{K}_{{{\max }_{2}}}}={{E}_{1}}-{{E}_{2}}=1.03eV\] \[{{V}_{1}}-{{V}_{2}}=1.03V\] \[(\because K=eV)\]
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