Lines in one spectral series can overlap lines in another.
(a) Use the Rydberg equation to see if the range of wavelengths in the n1 = 1 series overlaps the range in the n1 = 2 series.
(b) Use the Rydberg equation to see if the range of wavelengths in the n1 = 3 series overlaps the range in the n1 = 4 series.
(c) How many lines in the n1 = 4 series lie in the range of the n1 = 5 series?
(d) What does this overlap imply about the hydrogen spectrum at longer wavelengths?

Question

04-02-2020
Chemistry

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Lines in one spectral series can overlap lines in another.
(a) Use the Rydberg equation to see if the range of wavelengths in the n1 = 1 series overlaps the range in the n1 = 2 series.
(b) Use the Rydberg equation to see if the range of wavelengths in the n1 = 3 series overlaps the range in the n1 = 4 series.
(c) How many lines in the n1 = 4 series lie in the range of the n1 = 5 series?
(d) What does this overlap imply about the hydrogen spectrum at longer wavelengths?

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Answer:

(a) No overlap

(b) There is overlap

(c) Two

(d) See explanation below

Explanation:

1/λ = Rh (1/n₁² - 1/n₂² )

where λ is the wavelength of the transion, n₁ and n₂ are the principal energy levels ( n₁ < n₂ )

To solve this question, our strategy is to :

1. Calculate the longest wavelength for n₁ = 1, which corresponds to the transition with n₂ = 2.

2. Calculate the shortest wavelength for n₁ = 2, which corresponds to n₂ = infinity.

3. Compare the values to check if there is overlap

Lets plug the numbers to visualize this better:

Rydberg´s equation : 1/λ = 1.097 x 10⁷ /m x (1/n₁² - 1/n₂² )

For n₁ = 1, longest wavelength ( n₂ = 2 ) :

1/λ = 1.097 x 10⁷ /m x (1/1 ² - 1/2² ) = 8227.5/m

λ = 1/8227.5/m = 121 x 10⁻⁴ m x 1 x 10⁹ nm/m = 1.22 x 10² nm

For n₁ = 2, shortest wavelength ( n₂ = infinity ) :

1/λ = 1.097 x 10⁷ /m x (1/2 ² ) = 2.7 x 10⁶ /m

λ = 1/2.7 x 10⁶/m = 3.7 x 10⁻⁷ m x 1 x 10⁹ nm/m = 3.70 x 10² nm

There is no overlap between the n₁ = 1 and n₁ = 2 series ( there is no overlap 1.22 x 10² nm vs 3.70 x 10² nm )

(b) Repeat the same procedure as in part (a)

For n₁ = 3, longest wavelength ( n₂ = 4 ) :

1/λ = 1.097 x 10⁷ /m x (1/3 ² - 1/4² ) =5.33 x 10⁵/m

λ = 1/5.33 x 10⁵/m =1.88 x 10⁻⁶ m x 1 x 10⁹ nm/m = 1.88 x 10³ nm

For n₁ = 4, shortest wavelength ( n₂ = infinity ) :

1/λ = 1.097 x 10⁷ /m x (1/4 ² ) = 6.86 x 10⁵ /m

λ = 1/6.86 x 10⁵/m = 1.46 x10⁻⁶ m x 1 x 10⁹ nm/m = 1.46 x 10³ nm

There will be overlap

(c) Proceed as in the calculations above but now not only calculate for n₂ = 5 for n₁ = 4 but also a couple more and verify if there is overlap and count them.

For n₁ = 4 lets calculate n₂ = 5, 6, 7

1/λ = 1.097 x 10⁷ /m x (1/4 ² - 1/5² ) = 2.47 x 10⁵/m

λ = 1/2.47 x 10⁵/m = 4.05 x10⁻⁶ m x 1 x 10⁹ nm/m = 4.05 x 10³ nm

The same calculation is done for n₂ = 6 and 7, with the following results:

2.63 x 10³ nm, 2.17 x 10³ nm

Now the shortest wavelength in n₁ = 5 is:

1/λ = 1.097 x 10⁷ /m x (1/5² ) = 4.39 x 10⁵ / m

λ = 1/4.39 x 10⁵/m = 2.28 x10⁻⁶ m x 1 x 10⁹ nm/m = 2.28 x 10³ nm

There will be an overlap with 2 lines of n₁ = 4 (2.63 x 10³ nm, 2.17 x 10³ nm )

(d) The overlap tell us that the energy gap between energy levels becomes smaller as we could see from the calculations above. The spectra becomes confusing as there is more overlaps.