Question

The ratio of longest wavelength and the shortest wavelength observed in the five spectral series of emission spectrum of hydrogen is.

• A. $\frac{4}{3}$
• B. $\frac{525}{376}$
• C. 25
• D. $\frac{900}{11}$

Answer 1

Answer: (D)

$\frac{900}{11}$

Answer 2

Shortest wavelength comes from $\Rightarrow \frac{N}{10.38} = \left( \frac{1}{2} \right)^{5} \Rightarrow N = 10.38 \times \left( \frac{1}{2} \right)^{5}$to $n_{2} = 1$and longest wavelength comes from $n_{1} = 6$to $n_{2} = 5$in the given case. Hence $\frac{1}{\lambda_{\min}\left( \frac{1}{1^{2}} - \frac{1}{\infty^{2}} \right)}$

$\frac{1}{\lambda_{\max}\left( \frac{1}{5^{2}} - \frac{1}{6^{2}} \right)\left( \frac{36 - 25}{25 \times 36} \right)\frac{11}{900}}$

$\therefore\frac{\lambda_{\max}}{\lambda_{\min} = \frac{900}{11}}$.