Question

An electron undergoes a transition from an initial (ni) to a final (nf ) energy state. The energies of the ni and nf energy states are ?2.420

Answer 1

First let's find the difference in energy levels between the two states ni and nf.
That difference is the amount of energy contained in the photon emitted.
So the energy difference E = nf - ni = (-8.720 * 10**-20 J) - (-5.450 * 10**-19 J)
Making the exponents the same, -5.450 * 10**-19 = -54.50 * 10**-20 so we can
say E = (-8.720 - (-54.50)) * 10**-20 J = 45.78 * 10**-20 J or 4.578 * 10**-19 J
Now we know E, which is related to the wavelength w by: E = hc/w.
We want to find w, so w = hc/E. Since h and c are both constants, the product
hc is constant, (6.626 * 10**-34)(2.998 * 10**8) = 19.865 * 10**-26 Joule-meters.
So w = hc/E = (19.865 * 10**-26 J-m)/(4.578 * 10**-19 J) = 4.339 * 10**-7 meter.
The wavelength w is 4.339 * 10**-7 meter, or 433.9 nanometers (violet light).