Question

Problem 2. Calculate the ratio of wavelength of first spectral line of Lyman and Balmer series of hydrogen
spectrum.​

Answer 1

Answer:

5 : 27 .

Explanation:

According to Rydberg equation we have ,

$\displaystyle{\dfrac{1}{\lambda}=R\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right)}$

For the first line of Lyman series

$\displaystyle{n_1=1 \ and \ n_2=2}\\\\\displaystyle{\dfrac{1}{\lambda_l}=R\left( \dfrac{1}{1^2}-\dfrac{1}{2^2} \right) }\\\\\\\displaystyle{{\lambda_l}=R\left(\dfrac{3}{4}\right)}\\\\\\\displaystyle{\lambda_l=\dfrac{4}{3R}}$

For first line of Balmer series

$\displaystyle{n_1=2 \ and \ n_2=3}\\\\\displaystyle{\dfrac{1}{\lambda_b}=R\left(\dfrac{1}{2^2}-\dfrac{1}{3^2}\right)}\\\\\\\displaystyle{{\lambda_b}=R\left(\dfrac{5}{36}\right)}\\\\\\\displaystyle{\lambda_b=\dfrac{36}{5R}}$

Now taking ration of both ;

$\displaystyle{\dfrac{\lambda_l}{\lambda_b} =\left(\dfrac{\dfrac{4}{3R} }{\dfrac{36}{5R} }\right)}\\\\\\\displaystyle{\dfrac{\lambda_l}{\lambda_b} =\dfrac{5}{27}}$

Thus the ratio of Lyman and Balmer series of hydrogen  spectrum is 5 : 27 .

Answer 2

Explanation:

refer to the attachment