The hydrogen atom consists of a proton of mass *m _{p}*=1.7´10

in SI units.

If we confine ourselves to studying the relative motion of the two particles, and if we neglect any external forces, and if we treat the particles as spinless particles, then the Hamiltonian of the system is

.

The reduced mass *m* of the system is nearly the same
as the electron mass *m _{e}*, and the center of mass of the system is nearly
in the same place as the proton. We therefore often call the relative particle "the
electron" and the center of mass "the proton".

*.*

*H* is the Hamiltonian of a fictitious particle moving in a central potential.
We
already know that the eigenfunctions of *H* are of the form

,

where *u _{kl}(r)* satisfies the radial equation

with the condition *u _{kl}(0)=0*.

We define the effective potential .

In terms of the effective potential the radial equation becomes

.

As *r* approaches infinity, *V _{eff}(r)* goes to zero.
For every

For *E _{kl}<0*

In order to simplify the radial equation, we introduce the dimensionless quantity , where . The radial equation then becomes

.

We want to solve this equation for *E _{kl}<0* to find the energies of
the bound states of the hydrogen atom.

Defining

we have

.

We therefore try a solution of the form

.

*c _{kl} *satisfies the equation

.

We have

.

Therefore

.

We may write this as .

The coefficients *a _{q}* for each power

We have found a **recurrence relation** for
the coefficients *c _{q}*. Fixing

We have

.

As .

This implies that as *c _{kl}(r)* becomes proportional
to , since

.

The asymptotic behavior of *u _{kl}(r)* is then
dominated by which is physically
unacceptable.

In order to keep the wavefunction finite at infinity the power series defining must terminate.
For some integer *q=k* we need *c _{k}=0*. We then have

Since we require *c _{0}* to be non zero, the possible values for

*c _{kl}(r)* is a polynomial whose lowest
order term is

In terms of *c _{0}* the various coefficients

.

.

.

The **associate Laguerre polynomials** are
defined as

.

We therefore have

.

*c _{kl}*(

.

. Normalization requires .

.

We therefore have

.

.

For each *l* there exists an infinite number of possible energies, corresponding
to values of *k*=1, 2, 3, × × ×. Each of them is at
least *2l+*1 fold degenerate. This is the **essential**
degeneracy, since *E _{kl}* does not depend on

The possible eigenenergies therefore are .
Here *n* is called the **principal quantum
number**, *n* fixes the energy of the eigenstate. Given *n*, *l*
can take on *n* possible values . *n*
characterizes an **electron shell**, which contains *n*
**subshells** characterized by *l*. Each subshell
contains *2l+*1 distinct states.

The total degeneracy of the energy level *E _{n}* is for a hydrogen atom made from

- The wavefunctions of the hydrogen atom
- Probability densities (1)
- Probability Densities (2)
- The hydrogen atom

The ground state energy of the hydrogen atom is -*E _{I}*.

where

.

*a* is the **fine structure
constant**.

.

We also have

.

*R _{H}* is the

.

We denote the energy eigenstates of the hydrogen atom by { *|n,l,m>* }, where *n=k+l*.
The corresponding eigenfunctions are usually written as .

, where* k=n-l*, and* u _{kl}(r)* is given above. The number of radial nodes of

Letters of the alphabet are associated with various values of *l*.

l=0 | s |

l=1 | p |

l=2 | d |

l=3 | f |

l=4 | g |

Continue in alphabetic order.

The **subshells** are denoted by:

n |
l |
subshell |

n=1 |
l=0 |
1s |

n=2 |
l=0 |
2s |

n=2 |
l=1 |
2p |

etc.

The **principal shells** are denoted by:

n |
principal shell |

n=1 |
K-shell |

n=2 |
L-shell |

n=3 |
M-shell |

Continue in alphabetic order.

**Problem:**

In the interstellar medium electrons may recombine with protons to form
hydrogen atoms with high principal quantum numbers. A transition between
successive values of n gives rise to a recombination line.

(a) A radio recombination line occurs at 5.425978 *
10^{10} Hz for a n = 50 to n = 49 transition. Calculate the Rydberg constant for H.

(b) Compute the frequency of the H recombination line corresponding to
the transition n = 100 to n = 99.

(c) Assume the mean speed in part (b) is 10^{6} m/s. At what
frequency or frequencies would the recombination line be observable?

(d) Consider that radio recombination lines may be observed at either of two
facilities, the 11 meter telescope at Kitt Peak near Tuscon, Arizona, and the
1.2 meter radio telescope at Columbia University in New York. Relative larger
blocks of time are available on the smaller telescope, but its intrinsic noise
is moderately high. Where would you choose to map recombination radiation
emanating from an external galaxy. Discuss both technical and non technical
aspects of your choice.

Solution:

- Concepts:

The hydrogen atom, transitions between energy levels, the Doppler shift - Reasoning:

The wavelengths of the light emitted by excited hydrogen atoms can be found using the formula

1/λ = R_{H}(1/n^{2}- 1/n'^{2}). The observed wavelengths depend on the relative speed of the source and the observer. -
Details of the calculation:

(a) E_{n'}- E_{n}= hν = hcR_{H}(1/n^{2}- 1/n'^{2}).

5.425978 * 10^{10}= 2.99792458 * 10^{8}R_{H}(1/49^{2}- 1/50^{2}), R_{H}= 1.097373 * 10^{7 }m^{-1}.

(b) ν = cR_{H}(1/99^{2}- 1/100^{2}) = 6.679710 * 10^{9 }s^{-1}.

(c) The line is Doppler shifted.

ν' = ν[(1 + v/c)/(1 - v/c)]^{1/2}~ ν(1 + v/c) if v/c << 1.

Here v is the relative velocity of source and observer; v is positive if the source and the observer approach each other, and v is negative if the source and the receiver recede from each other.

-10^{6}/(3*10^{8}) < v/c < 10^{6}/(3*10^{8}), 6.657 * 10^{9 }s^{-1}< ν' < 6.702 * 10^{9 }s^{-1}.

(d) Here are some factors worth considering:

The amount of radiation gathered by a telescope is proportional to the square of its diameter D.

The smallest angle that can be resolved is approximately θ = λ/D.

(For each photon we have ΔxΔp_{x}~ h, Δx ~ D, ΔP_{x}~ h/D, θ = Δp_{x}/p = hc/(Dhν) = λ/D.)

If you assume that the radiation has a frequency of ~ 10^{-10}s^{-1}, then λ = c/ν = 3 cm. For the small telescope we therefore have θ = 0.03/1.2 while for the large telescope we have θ = 0.03/11. It takes (11/1.2)^{2}= 84 times as long to gather the same amount of radiation with the small telescope as with the large telescope.