A) \[2{{K}_{A}}={{K}_{B}}\]
B) \[{{K}_{A}}<{{K}_{B}}/2\]
C) \[{{K}_{A}}=2{{K}_{B}}\]
D) \[{{K}_{A}}={{K}_{B}}/2\]
Correct Answer: B
Solution :
\[\frac{hc}{\lambda }={{W}_{0}}+{{K}_{\max }}\Rightarrow \frac{hc}{{{\lambda }_{A}}}={{W}_{0}}+{{K}_{A}}\] (i) and \[\frac{hc}{{{\lambda }_{B}}}={{W}_{0}}+{{K}_{B}}\] (ii) Subtracting (i) from (ii), \[hc\left[ \frac{1}{{{\lambda }_{B}}}-\frac{1}{{{\lambda }_{A}}} \right]={{K}_{B}}-{{K}_{A}}\] \[\Rightarrow \]\[hc\left[ \frac{1}{{{\lambda }_{B}}}-\frac{1}{2{{\lambda }_{B}}} \right]={{K}_{B}}-{{K}_{A}}\] \[\Rightarrow \]\[\frac{hc}{2{{\lambda }_{B}}}={{K}_{B}}-{{K}_{A}}\] ?(iii) From (ii) and (iii), \[2{{K}_{B}}-2{{K}_{A}}={{W}_{0}}+{{K}_{B}}\] \[\Rightarrow \]\[{{K}_{B}}-2{{K}_{A}}={{W}_{0}}\] \[\Rightarrow \]\[{{K}_{A}}=\frac{{{K}_{B}}}{2}-\frac{{{W}_{0}}}{2}\]which gives \[{{K}_{A}}<\frac{{{K}_{B}}}{2}.\]
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