# 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)?

Dec 16, 2015

This is a bit long and probably there is a faster way but I tried this:

#### Explanation:

Ok; you know that when an electron jumps from an allowed orbit to another it absorbs/emits energy in form of a photon of energy $E = h \nu$ (where $h =$Planck's Constant and $\nu =$frequency).

Your "red" photon represents a transition between two orbits (of quantum numbers $n$ and $n + 1$) separated by a "energy" of:

${E}_{red} = h \cdot {\nu}_{red}$

but red means a wavelength: ${\lambda}_{red} = 656 n m$

so the "red" frequency will be: ${\nu}_{red} = \frac{c}{\lambda} _ \left(red\right) = \frac{3 \times {10}^{8}}{656 \times {10}^{-} 9} = 4.57 \times {10}^{14} H z$

And energy:

${E}_{red} = 6.63 \times {10}^{-} 34 \cdot 4.57 \times {10}^{14} = 3 \times {10}^{-} 19 J = 1.87 e V$
(where $1 e V = 1.6 \times {10}^{-} 19 J$)

Now we need to find the two orbits (their quantum numbers) from where and toward where the electron jumped:
We use Rydberg's Formula where we know that:
$\Delta E = {E}_{f} - {E}_{i} = 1.87 e V$
and also:
$\Delta E = - R \left(\frac{1}{n + 1} ^ 2 - \frac{1}{n} ^ 2\right)$
$1.87 = - 13.6 \left(\frac{1}{3} ^ 2 - \frac{1}{2} ^ 2\right)$
(I used Rydberg Constant in $e V$; $R = 13.6 e V$)
I found:
$1.87 = 1.88$ that works fine I think!

Hope it helps!