The correct relationship between wave numbers of spectral lines, in Balmer series of hydrogen atom having the least and highest wave - lengths, is given by

5 : 9

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9 : 5

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5 : 27

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27 : 5

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Solution

The correct option is B
9 : 5

In the Balmer series of hydrogen atom the highest wave length is that of first line and the least wave length is that of limiting line, thus
$\frac{{¯ ℧}_{\infty}}{{¯ ℧}_{1}} = \frac{\frac{1}{(2)^{2}} - \frac{1}{(\infty)^{2}}}{\frac{1}{(2)^{2}} - \frac{1}{(3)^{2}}} = \frac{\frac{1}{4} - 0}{(\frac{9 - 4}{36})} = \frac{1}{4} \times \frac{36}{5} = \frac{9}{5}$
or 9 : 5

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The correct relationship between wave numbers of spectral lines, in Balmer series of hydrogen atom having the least and highest wave - lengths, is given by

The correct option is B 9 : 5 In the Balmer series of hydrogen atom the highest wave length is that of first line and the least wave length is that of limiting line, thus ¯℧∞¯℧1=1(2)2−1(∞)21(2)2−1(3)2=14−0(9−436)=14×365=95 or 9 : 5