for the hydrogen atom, calculate the frequency and wavelength of light emitted for an electronic transition from n=5 to n=2

Question

I need to know how to solve this not just the answer

Answer 1

Hi Ariel!

 

For a hydrogen, or hydrogen-like, atom, the expression for the energy of an electron in the nth energy level, in units of electron-volts, is:

 

E_n = (-13.6 eV)/n²

 

If an electron moves from a higher energy level to a lower one, as in going from n=5 to n=2, it is a little like falling from a higher shelf to a lower one.  It loses an amount of energy equal to the difference in energy between the two levels.  That energy is emitted as light.

 

So the light, in this case, will have energy equal to:

 

E₅ - E₂ = [(-13.6 eV)/5²] - [(-13.6 eV)/2²] = 13.6((-1/25) - (-1/4)) = 13.6((1/4)-(1/25)) = 13.6(21/100) = 2.86 eV

 

[NOTE:  If you like, you can convert this to joules (J) using a conversion you can find on the internet or any class materials you may have]

 

The frequency of light, f, is related to its energy by the expression:

 

E = hf, where h = Planck's constant (= 4.14x10^-15 eV·s)

 

Knowing E and h, you can solve for f.

 

Then, the frequency of light and its wavelength, λ, are related by:

 

λ = c/f  where c = speed of light (2.99 x 10⁸ m/s)

 

This lets you solve for λ, once you have solved for f.

 

I hope this helps!  If you have more questions, or would like to check an answer, just let me know.