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Question: The ratio of minimum to maximum wavelength in Balmer series is-...

The ratio of minimum to maximum wavelength in Balmer series is-

A

59\frac{5}{9}

B

536\frac{5}{36}

C

14\frac{1}{4}

D

34\frac{3}{4}

Answer

59\frac{5}{9}

Explanation

Solution

For Balmer series 1λ\frac{1}{\lambda} = R [122−1n2]\left\lbrack \frac{1}{2^{2}} - \frac{1}{n^{2}} \right\rbrack

put n = , 1λmin⁡\frac{1}{\lambda_{\min}} = R4\frac{R}{4}

put n = 3, 1λmax⁡\frac{1}{\lambda_{\max}} = R [536]\left\lbrack \frac{5}{36} \right\rbrack

\ λmin⁡λmax⁡\frac{\lambda_{\min}}{\lambda_{\max}} = 5/361/4\frac{5/36}{1/4} = 59\frac{5}{9}