Q.26. In Balmer series, wavelength of the nd 2nd line is ‘λ1‘ and for Paschen series, wavelength of the 1st line is ‘λ2‘, then the ratio ‘λ1‘ to ‘λ2‘ is
A. 5:128
B. 5:81
C. 7:27
D. 9:132
Answer :- C. 7:27
Explanation :-
\frac{1}{\lambda}=R Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \text{For the Balmer series, } n_1=2 \text{The wavelength for } 2^{\text{nd}} \text{ line of the Balmer series is} \frac{1}{\lambda_1} = RZ^2\left(\frac{1}{2^2}-\frac{1}{4^2}\right) \frac{1}{\lambda_1} = RZ^2\left(\frac{1}{4}-\frac{1}{16}\right) \frac{1}{\lambda_1} = RZ^2\left(\frac{3}{16}\right) \Rightarrow \lambda_1=\left[\frac{16}{3}\right] \text{For the Paschen series, } n_1=3 \text{The wavelength for } 1^{\text{st}} \text{ line of the Paschen series is} \frac{1}{\lambda_2} = R Z^2\left(\frac{1}{3^2}-\frac{1}{4^2}\right) \frac{1}{\lambda_2} = R Z^2\left(\frac{1}{9}-\frac{1}{16}\right) \lambda_2 = \frac{144}{7} \frac{1}{\lambda_2} = R Z^2\left(\frac{7}{144}\right) \therefore \quad \frac{\lambda_1}{\lambda_2} = \frac{16}{3} \times \frac{7}{144}=\frac{7}{27}