Anonymous
Anonymous asked in Science & MathematicsChemistry · 9 years ago

# 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!

Relevance

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.