Answer:
The wavelength of next line in the series will be 397.05 nm
Explanation:
From Rydberg equation;
[tex]\frac{1}{\lambda} = R_H(\frac{1}{n_1^2} -\frac{1}{n_2^2})[/tex]
where;
λ is the wavelength
n lines in the series
RH is Rydberg constant = 1.097 x 10⁷ m⁻¹
Also at a given maximum wavelength, we can determine the first line n₁ in the series
[tex]\frac{1}{\lambda_{max}R_H} = \frac{1}{n_1^2} -\frac{1}{(n_1 +1)^2} \\\\[/tex]
[tex]\frac{1}{\lambda_{max}R_H} = \frac{2n_1+1}{n_1^2(n_1 +1)^2} \\\\\lambda_{max}R_H} = \frac{n_1^2(n_1 +1)^2}{2n_1+1}[/tex]
Given;
maximum wavelength = 656.46 nm
[tex]\lambda_{max}R_H} = \frac{n_1^2(n_1 +1)^2}{2n_1+1}\\\\656.46 *10^{-9}*1.097*10^7 = \frac{n_1^2(n_1 +1)^2}{2n_1+1}\\\\7.2 = \frac{n_1^2(n_1 +1)^2}{2n_1+1}[/tex]
Now, test for different values of n that will be equal to 7.2
let n₁ = 1
[tex]\frac{n_1^2(n_1 +1)^2}{2n_1+1} = \frac{(1)^2(1 +1)^2}{2(1)+1} = 1.3\\[/tex]
n₁(1) ≠ 7.2
Again, let n₁ = 2
[tex]\frac{n_1^2(n_1 +1)^2}{2n_1+1} = \frac{(2)^2(2 +1)^2}{2(2)+1} = 7.2\\[/tex]
∴ n₁(2) = 7.2
For the least wavelength given as 410.29 nm, n = ?
[tex]\frac{1}{\lambda} = R_H(\frac{1}{n_1^2} -\frac{1}{n_2^2})\\\\\frac{1}{410.29*10^{-9}} = 1.097*10^7(\frac{1}{2^2} -\frac{1}{n_2^2})\\\\\frac{1}{4.5} =\frac{1}{4} -\frac{1}{n_2^2}\\\\\frac{1}{n_2^2} =\frac{1}{4} -\frac{1}{4.5} \\\\\frac{1}{n_2^2} = 0.0277778\\\\n_2^2 = \frac{1}{0.0277778} \\\\n_2^2 = 36\\\\n_2= \sqrt{36}\ = 6[/tex]
next line in the series will be 7
The wavelength of next line in the series will be;
[tex]\frac{1}{\lambda} = R_H(\frac{1}{n_1^2} -\frac{1}{n_2^2})\\\\\frac{1}{\lambda} = 1.097*10^7(\frac{1}{2^2} -\frac{1}{7^2})\\\\\frac{1}{\lambda} = 1.097*10^7(0.22959)\\\\\frac{1}{\lambda} = 2518602.3\\\\\lambda = 397.05 \ nm[/tex]
Therefore, the wavelength of next line in the series will be 397.05 nm
