A) \[2{{k}_{1}}={{k}_{2}}\]
B) \[{{k}_{1}}=2{{k}_{2}}\]
C) \[{{k}_{1}}<{{k}_{2}}/2\]
D) \[{{k}_{1}}>2{{k}_{2}}\]
Correct Answer: C
Solution :
Here, \[{{K}_{1}}=\frac{hc}{{{\lambda }_{1}}}-W\] ???.(i) and \[{{K}_{2}}=\frac{hc}{{{\lambda }_{2}}}-W\] ??..(ii) Substituting \[{{\lambda }_{1}}=2{{\lambda }_{2}}\] in Eq. (i), we get \[{{K}_{1}}=\frac{hc}{2{{\lambda }_{2}}}-W\] \[{{K}_{1}}=\frac{1}{2}\left( \frac{hc}{{{\lambda }_{2}}} \right)-W\] \[=\frac{1}{2}({{K}_{2}}+W)-W\] \[{{K}_{1}}=\frac{{{K}_{2}}}{2}-\frac{W}{2}\] or \[{{K}_{1}}<\frac{{{K}_{2}}}{2}\]You need to login to perform this action.
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