# Atomic Structure Exercises?

What is the energy of a photon given off by one excited hydrogen atom as the electron moves from energy level 3 to energy level 2?

Any help would be greatly appreciated, if you could also please explain how you got the answer that would be amazing! Thank you!

### 1 Answer

- 9 years agoFavorite Answer
You can do this in two ways.

A. You can calculate the wavenumber (1/λ) of the emitted photon, using the equation …

1/λ = R∞ [1/(n2)^2 - 1/(n1)^2]

… where n1 and n2 are the principal quantum numbers of the energy levels involved in the transition, and R∞ is the Rydberg constant, 1.096 x 10^7 m^-1

Knowing the wavenumber, you can calculate the associated energy E using …

E = (hc) x (wavenumber)

... where h is Planck’s constant (6.626 x 10^-34 J.s) and c is the speed of light (3.00 x 10^8 m/s).

OR

B. You can go direct to the energy lost during the transition, using ...

E = Ry [1/(n2)^2 - 1/(n1)^2]

… where Ry is the Rydberg unit of energy, equal to 2.179 x 10^-18 J.

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Obviously, if you know the value of Ry, the second method is a little shorter.

E = Ry [1/(n2)^2 - 1/(n1)^2]

= (2.179 x 10^-18 J) [ (1/2^2) - (1/3^2)]

= (2.179 x 10^-18 J) [ 1/4 - 1/9]

= (2.179 x 10^-18 J) [0.2500 - 0.1111]

= (2.179 x 10^-18 J) [0.1389]

= 3.027 x 10^-19 J

The energy carried away by the photon is 3.027 x 10^-19 J.

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This photon from the n=3 ---> n=2 transition is the H-α photon, with a wavelength of 656 nm in the red region of the visible spectum.

I hope this is of some use to you.