Question

Consider the following portion of the energy-level diagram for hydrogen:

			n=4 –0.1361 × 10–18 J n=3 –0.2420 × 10–18 J n=2 –0.5445 × 10–18 J n=1 –2.178 × 10–18 J

In the hydrogen spectrum, what is the wavelength of light associated with the n = 2 to n = 1 electron transition? Please explain. The answer is: 1.22 × 10^–7 m.

Answer 1

To answer this question we need to understand what is this about first. "n" is the energy of an electron and it indicates which orbit it occupies. For example, if n=1, the electron is in the first orbit, closest to the nucleus. If n is 2, the electron is farther away and has a higher energy. And so on...

When the atom absorbs energy, electrons can jump from an orbit of a given enery to a higher orbit, e.g from first orbit to second. The electron then remains in that orbit until it relaxes to a lower enery state or a lower orbit and then energy is emitted.

This energy can be calculated with the difference between the energies in the provided list. In that case, let's compute Energy:

E = Efinal - Einitial = -2.18 x 10^-18J - (-0.544 x 10^-18J) = -1.636 x 10^-18 J.

Now we use the formula :

E = hc / lamdba

Where:

E = Energy (we use the absolute value)
h = Planck's constant which is 6.626*10^-34 J*s
c = Speed of light which is 3.00*10^8 m/s
Lambda = wavelength'

Finally,

Lambda = hc/e

Lamdba = (6.626*10^-34 J*s)(3.00*10^8m/s) / 1.635*10^-18 J = 1.22*10^-7 m