Question

For the Balmer series i.e., the atomic transitions where final state of the electron is n = 2, what is the longest and shortest wavelength possible? Is any of the frequency of Lyman series, which corresponds to transitions where electron ends up in n = 1 level, in the visible region?

Answer 1

The atomic transitions where final state of the electron is n = 2, atoms emit a series of lines in the visible part of the spectrum. This series is called the Balmer Series. Balmer examined the four visible lines in the spectrum of the hydrogen atom. Wave length of this series can be calculated by using the equation,

1/λ = R[1/2² - 1/n²],  where n = 3, 4, 5, ....

R is the Rydberg constant, whose value is 1.097 × 10⁷ m⁻¹ . The number n is just an integer; the above formula gives the longest wavelength, when n = 3, and gives each of the shorter wavelengths as n increases. From Balmer’s equation, it looks like when n gets bigger, the lines should start getting really close together. That’s exactly right; as n gets larger, 1 over n squared gets smaller, so there’s less and less difference between the consecutive lines. We can see that the series has a limit, that is, as n gets larger and larger, the wavelength gets closer and closer to one particular value. If n is infinity, then 1 over n squared is 0, and the corresponding wavelength is shortest. The individual lines in the Balmer series are given the names Alpha, Beta, Gamma, and Delta, and each corresponds to a n_i value of 3, 4, 5, and 6 respectively as shown in the figure below.

A picture from experimental demonstration is shown below.

The longest wavelength corresponds to the smallest energy difference between energy levels, which in this case will be between n = 2 and n = 3.

Wavelength for transition from n = 3

Thus the longest wavelength in the Balmer series is 656 nm. 

The shortest wavelength corresponds to the largest energy difference between energy levels, which in this case will be between n = 2 and n = ∞.

Wavelength for transition from n = ∞

Thus the shortest wavelength in the Balmer series is 365 nm. 

In order to see that whether any of the frequency of Lymen series is in the visible region, let’s calculate the largest and smallest frequencies of Laymen series. the wavelength of the Laymen series is given by

1/λ = R[1 - 1/n2], where n = 2, 3, 4, 5, ....

The longest wavelength corresponds to the smallest energy difference between energy levels, which in this case will be between n = 1 and n = 2.

Wavelength for transition from n = 2

Thus the longest wavelength in the Laymen series is 122 nm, and corresponding frequency will be,

The shortest wavelength corresponds to the largest energy difference between energy levels, which in this case will be between n = 1 and n = ∞.

Wavelength for transition from n = ∞

Thus the shortest wavelength in the Laymen series is 91.2 nm, and corresponding frequency will be,

From above we can see that frequency range of Laymen series is 24.7 × 10¹⁴ Hz to 32.9 × 10¹⁴. As frequency range of visible light is from 4 × 10¹⁴ to 8 × 10¹⁴, therefore no frequency of Laymen series is in the visible region.