A) \[{{K}_{1}}>\frac{{{K}_{2}}}{3}\]
B) \[{{K}_{1}}<\frac{{{K}_{2}}}{3}\]
C) \[{{K}_{1}}=3{{K}_{2}}\]
D) \[{{K}_{2}}=3{{K}_{1}}\]
Correct Answer: B
Solution :
Energy of each photon is \[E=\frac{hc}{\lambda }-\frac{hc}{{{\lambda }_{0}}}\] Case \[{{K}_{1}}=\frac{hc}{{{\lambda }_{1}}}-\frac{hc}{{{\lambda }_{0}}}\] \[{{K}_{1}}=\frac{hc}{3{{\lambda }_{2}}}-\frac{hc}{{{\lambda }_{0}}}\] Case II ?(1) \[{{K}_{2}}=\frac{hc}{{{\lambda }_{2}}}-\frac{hc}{{{\lambda }_{0}}}\] ?(2) From Eq. (1) \[3{{K}_{1}}=\frac{hc}{{{\lambda }_{2}}}-\frac{3hc}{{{\lambda }_{0}}}\] \[3{{K}_{1}}={{K}_{2}}+\frac{hc}{{{\lambda }_{0}}}-\frac{3hc}{{{\lambda }_{0}}}\] \[3{{K}_{1}}={{K}_{2}}-\frac{2hc}{{{\lambda }_{0}}}\] \[3{{K}_{1}}<{{K}_{2}}\] \[{{K}_{1}}<\frac{{{K}_{2}}}{3}\]You need to login to perform this action.
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