Answer:
(b)>(c)>(a) >(d)
Explanation:
We know that from Rydberg´s equation:
1/ λ = Rh x (1/n₁² - 1/n₂² )
where n₁ and n₂ are the principal quantum numbers involved in the transition, and n₁ < n₂.
Therefore the wavelength will be given by taking the reciprocal of this equation:
λ = 1 / [Rh x (1/n₁² - 1/n₂² ) ]
So lets calculate the wavelengths for the transitions in this question expressed in terms of the constant Rh
(a ) λ = 1 / [Rh x (1/2² ) ] = 4/ Rh
(b ) λ = 1 / [Rh x (1/4² - 1/20² ) ] = 16.7 / Rh
(c) λ = 1 / [Rh x (1/3² - 1/10² ) ] = 9.9 / Rh
(d) λ = 1 / [Rh x (1/1² - 1/2² ) ] = 1.33 / Rh
Therefore in decreasing wavelength is (b)>(c)>(a) >(d)
as shown in these calculations, be careful with this type of question, since one might erroneously think that the transition for example as in (a) will have a shorter wavelength than (d) which is not the case as shown here. One must use Rydbergs formula.
Arranging the H atom electron transitions in decreasing order ;
B ---> C ----> A -----> D
Determine the wavelength of the photon absorbed by the H atom electron
we will apply Rydberg's equation
Rydberg equation = 1/ λ = Rh * (1/n₁² - 1/n₂² ) --- ( 1 )
n = quantum numbers
λ = wavelength
∴ λ = 1 / [Rh * (1/n₁² - 1/n₂² ) ] ---- ( 2 )
a) For n = 2 to n = [infinity]
λ = 1 / [ Rh * ( 1 / 2² - ∞ ) = 4 / Rh
b) For n = 4 to n= 20
λ = 1 / [ Rh * ( 1 / 4² - 1 / 20² ) = 16.7 Rh ( highest wavelength )
c) For n = 3 to n = 10
λ = 1 / [ Rh * ( 1 / 3² - 1 / 10² ) = 9.9 Rh
d) For n = 2 to n = 1
λ = 1 / [ Rh * ( 1 / 2² - 1 / 1² ) = 1.33 Rh ( Lowest wavelength )
Therefore arranging the H atom electron transitions in order of decreasing wavelength will be; B ---> C ----> A -----> D.
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