A) \[{{n}_{4}}\to {{n}_{1}}\]
B) \[{{n}_{2}}\to {{n}_{1}}\]
C) \[{{n}_{4}}\to {{n}_{2}}\]
D) \[{{n}_{3}}\to {{n}_{1}}\]
Correct Answer: A
Solution :
[a] \[{{n}_{4}}\to {{n}_{1}}\] According to Bohr's theory, \[\Delta E=\,hv=\frac{hc}{\lambda }\] \[\Rightarrow \lambda =\frac{hc}{\Delta E}\] \[i.e.\,\lambda \propto \,\frac{1}{\Delta \,E}\] \[\therefore \Delta E={{E}_{4}}-{{E}_{1}}\] will be maximum and hence wavelength will be minimum from \[{{n}_{4}}\to {{n}_{1}}\]
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