Answered

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light of wavelength 121.6 nm is emitted by a hydrogen atom. what are the (a) higher quantum number and (b) lower quantum number of the transition producing this emission? (c) what is the name of the series that includes the transition?

Sagot :

(a) higher quantum number , n2 = 2

(b) lower quantum number of the transition producing this emission, n1 = 1

(c)  the name of the series that includes the transition, n = 2 to n = 1

Given that :

wavelength of light = 121.6 nm

wave number = 1 / λ

                       = 1 / 121.6 × 10⁻⁹ m

                      = 8.22 × 10⁶ m⁻¹

the Rydberg equation is given as :

1 / λ = R ( 1 / n1² - 1 / n2² )

n1 =

n2 = 2, 3, 4...

1 / λ = R ( 1 / n1² - 1 / n2² )

1 / λ  = 1.097 × 10⁷ ( 1 / 1 - 1 / 2² )

1 / λ = 8.22 × 10⁶ m⁻¹

it is clear from the above equation , n1 = 1 and n2 = 2

To learn more about wavelength here

https://brainly.com/question/20533910

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