A) 2 : 3
B) 3 : 2
C) 1 : 5
D) 5 : 1
Correct Answer: A
Solution :
The energy of photon \[E=hv=\frac{hc}{\lambda }\] \[E\propto \frac{1}{\lambda }\] \[\therefore \] \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{{{\lambda }_{2}}}{{{\lambda }_{1}}}=\frac{4000}{6000}=\frac{2}{3}\]
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