Answer :
(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
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