Question
Question: Given the value of Rydberg constant is, \({{10}^{7}}{{m}^{-1}}\)the waves number of the last line of...
Given the value of Rydberg constant is, 107m−1the waves number of the last line of the Balmer series in hydrogen spectrum will be:
(A). 0.025×104m−1
(B). 0.5×107m−1
(C). 0.25×107m−1
(D). 2.5×107m−1
Solution
Hint: Reciprocal of wavelength is wave number. Wave number gives number of lines per distance. The Hydrogen spectrum has four series. Namely: Lyman series, Balmer series, Paschen series, Brackett series, Pfund series. Use the formula of wave number.
Complete step by step answer:
We know that the Balmer series is the second line of the hydrogen spectrum.
So we have formula of wave number:
ν=λ1=RZ2(n211−n221) --------(1)
Where,
ν = wave number
λ = wavelength
R = Rydberg constant
Z = Atomic number
n1=First transition state
n2= next transition state
Given that it is a Balmer series. Therefore in Balmer series:
n1=2 and n2= infinity (∞)
R = Rydberg constant = 10−7
It is a hydrogen atom, therefore the atomic number must be equal to one.
Z = 1
Put value of R,Z,n1,n2 in equation (1)
We get,
ν=λ1=107×12(221−∞21)