Question

A series of lines involving a common level in the spectrum of atomic hydrogen lies at 656.46 nm, 486.27 nm, 434.17 nm, and 410.29 nm. What is the wavelength of the next line in the series

Answer 1

Answer:

The wavelength of next line in the series will be 397.05 nm

Explanation:

From Rydberg equation;

$\frac{1}{\lambda} = R_H(\frac{1}{n_1^2} -\frac{1}{n_2^2})$

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

$\frac{1}{\lambda_{max}R_H} = \frac{1}{n_1^2} -\frac{1}{(n_1 +1)^2}$

$\frac{1}{\lambda_{max}R_H} = \frac{2n_1+1}{n_1^2(n_1 +1)^2} \\\\\lambda_{max}R_H} = \frac{n_1^2(n_1 +1)^2}{2n_1+1}$

Given;

maximum wavelength = 656.46 nm

$\lambda_{max}R_H} = \frac{n_1^2(n_1 +1)^2}{2n_1+1}\\\\656.46 *10^{-9}*1.097*10^7 = \frac{n_1^2(n_1 +1)^2}{2n_1+1}\\\\7.2 = \frac{n_1^2(n_1 +1)^2}{2n_1+1}$

Now, test for different values of n that will be equal to 7.2

let n₁ = 1

$\frac{n_1^2(n_1 +1)^2}{2n_1+1} = \frac{(1)^2(1 +1)^2}{2(1)+1} = 1.3$

n₁(1) ≠ 7.2

Again, let n₁ = 2

$\frac{n_1^2(n_1 +1)^2}{2n_1+1} = \frac{(2)^2(2 +1)^2}{2(2)+1} = 7.2$

∴ n₁(2) = 7.2

For the least wavelength given as 410.29 nm, n = ?

$\frac{1}{\lambda} = R_H(\frac{1}{n_1^2} -\frac{1}{n_2^2})\\\\\frac{1}{410.29*10^{-9}} = 1.097*10^7(\frac{1}{2^2} -\frac{1}{n_2^2})\\\\\frac{1}{4.5} =\frac{1}{4} -\frac{1}{n_2^2}\\\\\frac{1}{n_2^2} =\frac{1}{4} -\frac{1}{4.5} \\\\\frac{1}{n_2^2} = 0.0277778\\\\n_2^2 = \frac{1}{0.0277778} \\\\n_2^2 = 36\\\\n_2= \sqrt{36}\ = 6$

next line in the series will be 7

The wavelength of next line in the series will be;

$\frac{1}{\lambda} = R_H(\frac{1}{n_1^2} -\frac{1}{n_2^2})\\\\\frac{1}{\lambda} = 1.097*10^7(\frac{1}{2^2} -\frac{1}{7^2})\\\\\frac{1}{\lambda} = 1.097*10^7(0.22959)\\\\\frac{1}{\lambda} = 2518602.3\\\\\lambda = 397.05 \ nm$

Therefore, the wavelength of next line in the series will be 397.05 nm