A) \[{{\lambda }_{1}}=\sqrt{{{\lambda }_{2}}}\]
B) \[{{\lambda }_{1}}<{{\lambda }_{2}}\]
C) \[{{\lambda }_{1}}={{\lambda }_{2}}\]
D) \[{{\lambda }_{1}}>{{\lambda }_{2}}\]
Correct Answer: D
Solution :
\[\because {{V}_{0}}=\left( \frac{h}{e} \right)\,\nu -\left( \frac{{{W}_{0}}}{e} \right)\]. From the graph \[{{V}_{2}}>{{V}_{1}}\] Þ \[\frac{h{{\nu }_{2}}}{e}-\frac{{{W}_{0}}}{e}>\frac{h{{\nu }_{1}}}{e}-\frac{{{W}_{0}}}{e}\] Þ \[{{\nu }_{2}}>{{\nu }_{1}}\] Þ \[{{\lambda }_{1}}>{{\lambda }_{2}}\] (as \[\lambda \propto \frac{1}{\nu }\])
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