# Using Rydberg's equation for the energy levels (E = -R/n^2) and the wavelength and the c=(lambda)v, E=hv equations, SHOW/ PROVE how E(red band in H-spectrum, lambda = 656 nm) = (Delta or change/...

# Using Rydberg's equation for the energy levels (E = -R/n^2) and the wavelength and the c=(lambda)v, E=hv equations, SHOW/ PROVE how E(red band in H-spectrum, lambda = 656 nm) = (Delta or change/ difference) E (3 --> 2)?

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### 1 Answer

We need to prove that the light emitted when an electron in Hydrogen atom changes its energy level from n = 3 to n = 2 has the wavelength of 656 nm.

The change of energy between the levels 3 and 2 is

`DeltaE_(32) = E_2 - E_3 = -E_0/2^2 -(-E_0/3^2) = -E_0/4+ E_0/9 = -(5E_0)/36` , where

`E_0 = 13.6 eV`

So, the energy released in transition from level with n = 3 to the level with n = 2 is

`|DeltaE| = (5*13.6)/36 eV =1.89 eV`

The frequency of the emitted light wave can be found from

`|DeltaE| = h*nu` , where h is the Planck's constant: `h=4.12*10^(-15) eV*s` .

From here

`nu = |DeltaE|/h = 1.89/(4.25*10^(-15)) = 4.6*10^14 1/s` .

Then, the wavelength of this light wave is

`lambda = c/nu = (3*10^8 m/s)/(4.6*10^14) = 6.54*10^(-7) m = 654 nm`

Taking into account the approximation from rounding the results during calculations, this shows that the wavelength of the light emitted from the transition from level 3 to level e is `lambda = 656 nm` .

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