38.6.2 Hydrogen Atom Energy Levels

Now we can make progress quickly. Knowing the possible radii, we can return to Equation 38.28 and find the possible electron speeds to be

v_{n} = \frac{n ℏ}{m r_{n}} = \frac{1}{n} \frac{ℏ}{m a_{B}} = \frac{v_{1}}{n} n = 1, 2, 3, \dots

(38.31)

where $v_{1} = ℏ / m a_{B} = 2.19 \times 10^{6} m / s$ is the electron’s speed in the $n = 1$ orbit. The speed decreases as $n$ increases.

Finally, we can determine the energies of the stationary states. From Equation 38.22 for the energy, with Equations 38.30 and 38.31 for $r$ and $v$ , we have

E_{n} = \frac{1}{2} m v_{n}^{2} - \frac{e^{2}}{4 π ϵ_{0} r_{n}} = \frac{1}{2} m (\frac{ℏ^{2}}{m^{2} a_{B}^{2} n^{2}}) - \frac{e^{2}}{4 π ϵ_{0} n^{2} a_{B}}

(38.32)

As a homework problem, you can show that this rather messy expression simplifies to

E_{n} = - \frac{1}{n^{2}} (\frac{1}{4 π ϵ_{0}} \frac{e^{2}}{2 a_{B}})

(38.33)

The expression in parentheses is easily evaluated, giving

\frac{1}{4 π ϵ_{0}} \frac{e^{2}}{2 a_{B}} = 13.60 eV

We can then write the energy levels of the stationary states of the hydrogen atom as

E_{n} = - \frac{13.60 eV}{n^{2}} n = 1, 2, 3, \dots

(38.34)

This has been a lot of math, so we need to see where we are and what we have learned. TABLE 38.2 shows values of $r_{n}, v_{n}$ , and $E_{n}$ evaluated for quantum number values $n = 1$ to 5. We do indeed seem to have discovered stationary states of the hydrogen atom. Each state, characterized by its quantum number $n$ , has a unique radius, speed, and energy. These are displayed graphically in FIGURE 38.19, in which the orbits are drawn to scale. Notice how the atom’s diameter increases very rapidly as $n$ increases. At the same time, the electron’s speed decreases.

TABLE 38.2 Radii, speeds, and energies for the first five states of the Bohr hydrogen atom

$n$	$r_{n} (nm)$	$v_{n} (m / s)$	$E_{n} (eV)$
1	0.053	$2.19 \times 10^{6}$	$- 13.60$
2	0.212	$1.09 \times 10^{6}$	$- 3.40$
3	0.476	$0.73 \times 10^{6}$	$- 1.51$
4	0.846	$0.55 \times 10^{6}$	$- 0.85$
5	1.322	$0.44 \times 10^{6}$	$- 0.54$

The figure illustrates the first four stationary states, or allowed orbits, of the Bohr hydrogen atom.

FIGURE 38.19 Full Alternative Text

EXAMPLE 38.11 Stationary states of the hydrogen atom

Can an electron in a hydrogen atom have a speed of $3.60 \times 10^{5} m / s ?$ If so, what are its energy and the radius of its orbit? What about a speed of $3.65 \times 10^{5} m / s ?$

SOLVE To be in a stationary state, the electron must have speed

v_{n} = \frac{v_{1}}{n} = \frac{2.19 \times 10^{6} m/s}{n}

where $n$ is an integer. A speed of $3.60 \times 10^{5} m / s$ would require quantum number

n = \frac{2.19 \times 10^{6} m/s}{3.60 \times 10^{5} m/s} = 6.08

This is not an integer, so the electron can not have this speed. But if $v = 3.65 \times 10^{5} m / s$ , then

n = \frac{2.19 \times 10^{6} m/s}{3.65 \times 10^{5} m/s} = 6

This is the speed of an electron in the $n = 6$ excited state. An electron in this state has energy

E_{6} = - \frac{13.60 eV}{6^{2}} = - 0.38 eV

and the radius of its orbit is

r_{6} = 6^{2} (5.29 \times 10^{- 11} m) = 1.90 \times 10^{- 9} m = 1.90 nm