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Atomic hydrogen produces several series of spectral lines. Each series fits the Rydberg equation with its own particular n₁ value. Calculate the value of n₁ (by trial and error if necessary) that would produce a series of lines in which:
(b) The lowest energy line has a wavelength of 7460 nm.

Respuesta :

The lowest energy line in a sequence of lines with a 7460 nm wavelength would result from a value of [tex]n_{1}[/tex] = 5.

Calculation of value of n₁ :

The formula that connects the Rydberg constant and the [tex]n_{1}[/tex] values to the wavelength of emissions is

(1/λ)  [tex]= R [(1/(n_{1} ^2)] - [1/(n_{2} ^2)][/tex]

Where,

λ = wavelength,

R = (10.972 × 10^6)/m,

[tex]n_{2}[/tex] = ∞ (since they have already been released from the atom)

λ = 7460 nm = (7.46 × [tex]10^{-6}[/tex])m

[1/(7.46 × [tex]10^{-6}[/tex])] = (10.972 × 10^6)[tex][(1/(n_{1} ^2)) - (1/(n_{2} ^2)][/tex]

lowest energy line,[tex]n_{2}[/tex] = n₁ + 1

[tex](n_{1} ^2)[(n_{1} +1)^2)]/(2n_{1} +1)[/tex] = (7.46 × 10^-6) × (10.972 × 10^6) = 81.85

By solving quadratic eqn,

n₁ = 5

Hence, the lowest energy line in a sequence of lines with a 7460 nm wavelength would result from a value of [tex]n_{1}[/tex] = 5.

Learn more about Rydberg equation here:

https://brainly.com/question/14649095

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