A) n = 1 to n = 2
B) n = 2 to n = 4
C) n= 3ton= 6
D) n = 1 to n = \[\infty \]
Correct Answer: D
Solution :
\[\Delta \Epsilon =cons\tan t\]\[\left[ \frac{1}{{{n}_{1}}^{2}}-\frac{1}{n_{2}^{2}} \right],\] Transition with maximum valu e of \[{{n}_{2}}\]and least value of \[{{n}_{1}}\] will absorb greatest quantity of energy \[(n=1\,to\,n=\infty )\]
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