Question

the ratio of minimum to maximum wavelength in balmer series

Answer 1

Hope it helps you... If u have any query regarding the solution then do ask me....

Answer 2

Answer:

The ratio of minimum to maximum wavelength in balmer series is $\bold{{\frac{\lambda_{\min }}{\lambda_{\max }}=5 : 9}\end{array}}$

Explanation:

The Balmer series have their specific wavelength attained through their own series equation.  

As we know that, for the Balmer series, $\bold{n_{1}}$ = 2.

Thereby, considering the minimum wavelength, we have $\bold{n_{2}}$ = infinity

Similarly, considering the maximum wavelength, we have $\bold{n_{2}}$ = 3.

So from the Balmer series equation we have – $\begin{aligned} \frac{1}{\lambda_{\min }} &=R_{n}\left[\left(\frac{1}{n_{1}^{2}}\right)-\left(\frac{1}{n_{2}^{2}}\right)\right] \\ \frac{1}{\lambda_{\min }} &=R_{n}\left[\left(\frac{1}{2^{2}}\right)-\left(\frac{1}{\infty^{2}}\right)\right] \\ \frac{1}{\lambda_{\min }} &=R_{n}\left(\left(\frac{1}{4}\right)-\left(\frac{1}{\infty}\right)\right) \\ \frac{1}{\lambda_{\min }} &=R_{n}\left[\frac{1}{4}\right] \\ \lambda_{\min } &=\frac{4}{R_{n}} \\ \lambda_{\min } &=\frac{4}{R_{n}} \end{aligned}$

$\begin{array}{l}{\text { And, } \frac{1}{\lambda_{\max }}=R_{n}\left[\left(\frac{1}{n_{1}^{2}}\right)-\left(\frac{1}{n_{2}^{2}}\right)\right]} \\ {\frac{1}{\lambda \max }=R_{n}\left[\left(\frac{1}{2^{2}}\right)-\left(\frac{1}{3^{2}}\right)\right]} \\ {\frac{1}{\lambda \max }=R_{n}\left[\left(\frac{1}{4}\right)-\left(\frac{1}{9}\right)\right]} \\ {\frac{1}{\lambda_{\max }}=R_{n}\left(\frac{5}{36}\right)} \\ {\lambda_{\max }=\frac{36}{5} R_{n}}\end{array}$

Thereby,$\begin{aligned} \frac{\lambda_{\min }}{\lambda_{\max }} &=\frac{\frac{4}{R_{n}}}{\frac{36}{5 R_{n}}} \\ \frac{\lambda_{\min }}{\lambda_{\max }} &=\frac{4}{R_{n}} \times \frac{5 R_{n}}{36} \end{aligned}  \begin{array}{l}{\frac{\lambda_{\min }}{\lambda_{\max }}=\frac{5}{9}} \\ {\frac{\lambda_{\min }}{\lambda_{\max }}=5 : 9}\end{array}$