Solution

Substituting for R and inverting gives A = 656.47 nm

This is the wavelength as measured in a vacuum. We generally refer to the wavelength in air, since that is how we measure it at a telescope. The wavelength in air is that in vacuum divided by the index of refraction of air, 1.000 29, giving 656.28 nm. (It is interesting to note that when spec-troscopists tabulate wavelengths, those longer than 200 nm are given as they would be in air, since that is how they usually will be measured. Radiation with wavelengths less than 200 nm doesn't penetrate through air, and its wavelengths are usually measured in a vacuum, so the vacuum values are tabulated.)

3.3.1 The Bohr atom

The next advancement was by the Danish physicist Neils Bohr who tried to understand hydrogen (the simplest atom), illustrated in Fig. 3.4. He postulated the existence of certain stationary states. If the electron is orbiting in one of these states, the atom is stable. Each of these states has a particular energy. We can let the energy of the nth state be En and the energy of the mth state be Em. For definiteness, let En > Em.

Under the right conditions, transitions between states can take place. If the electron is in the higher energy state, it can drop down to the lower energy state, as long as a photon is emitted with an energy equal to the energy difference between

4th Orbit

4th Orbit

the two states. If the frequency of the photon is v, We put this into equation (3.2) to find the total this means that energy in terms of r:

If the electron is in the lower energy state, it The minus sign indicates that the total energy is can make a transition to the upper state if the negative. To see what this means, remember that atom absorbs a photon with exactly the right we have defined the potential energy such that it energy. This explanation incorporated Einstein's is zero when the electron and proton are infi-idea of photons. nitely far apart. The electron and proton being far

Bohr pointed out that one could calculate the apart with no motion is the minimal condition energies of the allowed states by assuming that for the electron being free of the proton. So, if the the angular momentum J of the orbiting elec- electron is barely free of the proton, the total trons is quantized in integer multiples of h/2w. energy would be zero. So, if the total energy is The combination h/2w appears so often we give it negative, as in equation (3.4), then we must add its own symbol, "h (spoken as 'h-bar'). We apply energy [(1/2)e2/r] if we want to bring it up to zero, this to a hydrogen atom, with an electron, with which would free the electron. So the negative charge -e, orbiting a distance r from a nucleus energy means that the electron is not free. In this with charge +e. We assume that the nucleus is case we say that the system is bound. much more massive than the electron, so we can We still have to find the allowed values of r. ignore the small motion of the nucleus, since The angular momentum is J = mvr. The quantiza-both the nucleus and electron orbit their com- tion condition becomes mon center of mass.

We first look at the kinetic energy

Solving for v gives

The potential energy, relative to the potential energy being zero when the electron is infinitely Squaring and multiplying by m gives far from the nucleus, is given by mv2 = n2h2/4w2 mr2

(By writing the potential energy in this form, e2/r = n2h2/4w2 mr2 rather than with a factor of 1/4we0, we are using cgs units. This means that charges are expressed We now solve for r, giving the radius of the nth in electrostatic units, esu, with the charge on the orbit:

electron being 4.8 X 10-10 esu.) The total energy rn = n2h2/4w2me2 (3.5)

is the sum of kinetic and potential:

Substituting into equation (3.4) gives the energy E = (1/2)mv2 - e2/r (3.2) of the nth state:

We can relate v and r by noting that the elec- En = -(1/2)e4m(4w2)/n2h2 (3.6)

trical force between the electron and the nucleus, 2

e2/r2, must provide the acceleration to keep the Note that this has the 1/n dependence that we electron in the circular orbit, v2/r. This tells us would expect from the Balmer formula.

that One modification that we should make is to account for the motion of the nucleus (since it is mv2/r = e2/r2 not infinitely massive). We should replace the

„ u. , . i ■ j „ , mass of the electron, m, in equations (3.5) and

Multiplying both sides of the equation by r i \ /

(3.6) by the reduced mass of the electron and pro-

gives ton. The reduced mass, mr, is defined such that mv2 = e2/r (3.3) the motion of the electron, as viewed from the

(moving) proton, is as if the proton were fixed and the electron's mass is reduced to mr. An expression for mr is (see Problem 3.2)

0.9995 me where me and mp are the masses of the electron and proton, respectively.

Example 3.2 Hydrogen atom energy Compute the energy of the lowest (ground) energy level in a hydrogen atom. Also, find the radius of the orbit of the electron in that state.

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