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3 years ago

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 410.2

julsineya [31]

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;

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