The atomic mass is the mass of an atom. The atomic mass or relative isotopic mass refers to the mass of a single particle, and therefore is tied to a certain specific isotope of an element. The atomic mass is carried by the atomic nucleus, which occupies only about 10 -12 of the total volume of the atom or less, but it contains all the positive. For example, the mass of one unbound atom of the common hydrogen isotope (hydrogen-1, protium) is 1.007 825 032 241 (94) Da, the mass of one free neutron is 1.008 664 915 95 (49) Da, and the mass of one hydrogen-2 (deuterium) atom is 2.014 101 778 114 (122) Da.
The hydrogen atom consists of a proton of mass mp=1.7´10-27kg and charge qe=1.6´10-19C and an electron of mass me=9.1´10-31kg and charge -qe. The dominant part of the interaction between the two particles is the electrostatic interaction. The electrostatic potential energy is
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 mof the system is nearly the same as the electron mass me, 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'.
The eigenvalues and eigenfunctions of H(r,p)
.
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 ukl(r) satisfies the radial equation
with the condition ukl(0)=0.
We define the effective potential .
In terms of the effective potential the radial equation becomes
.
As r approaches infinity, Veff(r) goes to zero. For every Ekl>0 there therefore exists a solution ukl(r) with an oscillatory behavior at infinity. These solutions represent unbound states.
For Ekl<0ukl(r) will be proportional to as r approaches infinity. The real positive exponential solution is physically unacceptable. As r goes to zero, ukl(r) becomes proportional to rl+1. Only for certain discrete values of Ekl is it possible to properly match these two asymptotic solutions.
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 Ekl<0 to find the energies of the bound states of the hydrogen atom.
Defining
we have
.
We therefore try a solution of the form
.
cklsatisfies the equation
.
We have
.
Therefore
.
We may write this as .
The coefficients aq for each power q of r must vanish for this to hold for all r. For the power q+l-1 we therefore must have . .
We have found a recurrence relation for the coefficients cq. Fixing c0 allows us to calculate all cq.
We have
.
As .
This implies that as ckl(r) becomes proportional to , since
.
The asymptotic behavior of ukl(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 ck=0. We then have ck(q>k)=0. For ck=0 we need .
Since we require c0 to be non zero, the possible values for k are k=1, 2, 3, × × ×. For a given l the possible values of Ekl then are .
ckl(r) is a polynomial whose lowest order term is rl+1and whose highest order term is rk+l.
In terms of c0 the various coefficients cq are calculated using the recurrence relations
.
.
.
Mass Of The Hydrogen Atom
The associate Laguerre polynomials are defined as
.
We therefore have
.
ckl(r) is proportional to a polynomial of order k-1 and therefore has k-1 radial nodes.
.
State The Mass Of One Atom Of Hydrogen In Kg
. Normalization requires .
.
We therefore have
.
.
Energy levels
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 Ekl does not depend on m. Accidental degeneracies also exist. For the hydrogen atom we find that Ekl is only a function of k+l, and we define k+l=n for the hydrogen atom.
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 nsubshells characterized by l. Each subshell contains 2l+1 distinct states.
The total degeneracy of the energy level En is for a hydrogen atom made from spinless particles.
Links:Definitions and notations
The ground state energy of the hydrogen atom is -EI.
where
.
a is the fine structure constant.
.
We also have
.
RH is the Rydberg constant.
.
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, andukl(r) is given above. The number of radial nodes of Rnl(r) is n-l-1.
Spectroscopic notation
Letters of the alphabet are associated with various values of l.
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:
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 * 1010 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 106 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 = RH(1/n2 - 1/n'2). The observed wavelengths depend on the relative speed of the source and the observer. Details of the calculation:
(a) En' - En = hν = hcRH(1/n2 - 1/n'2).
5.425978 * 1010 = 2.99792458 * 108RH(1/492 - 1/502), RH = 1.097373 * 107 m-1.
(b) ν = cRH(1/992 - 1/1002) = 6.679710 * 109 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.
-106/(3*108) < v/c < 106/(3*108), 6.657 * 109 s-1 < ν' < 6.702 * 109 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 ΔxΔpx ~ h, Δx ~ D, ΔPx ~ h/D, θ = Δpx/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.
So then, why isn't the atomic mass of Hydrogen exactly 1?
If you check a periodic table, you'll see that Hydrogen actually has a mass of 1.00794. If hydrogen is the lightest of all substances, then why not give it a mass of exactly 1 on our relative mass scale?
There are three reasons:
- First, atoms have isotopes, and these isotopes do not all have the same mass. The mass of the atoms in nature - what we use as the atomic mass - is a weighted average of all these different isotopes.
Here are the exact atomic masses and abundances of an atom with two imaginary stable isotopes.
To 4 significant digits, what would be the calculated atomic mass of naturally occurring X? |
- The second reason is historical. Once upon a time, way back before 1961, there actually were two sets of atomic masses (though everybody called them atomic weights then). One scale was used by physicists; the other by chemists. Both were based on weights compared to Oxygen, rather than Hydrogen. Oxygen was used because it combines with a lot of things to form oxides. This made it a better choice as a standard because of the ease of chemical analysis. Oxygen was set to have an atomic mass of 16, which was just about 16 times as heavy as Hydrogen being 1. Unfortunately, Chemists picked naturally occurring Oxygen, which is a mixture of isotopes of Oxygen-16, Oxygen-17, and Oxygen-18. After all when you made an oxide of an element you would do so in naturally occurring oxygen. Physicists picked the pure isotope Oxygen-16, because they tended to make their measurements on the basis of mass spectrometry.
Though the ratio of any two atom's masses was the same on either scale, it was horribly confusing, so in 1961, a compromise was reached. Instead of using either Hydrogen, or Oxygen as the standard, the isotope of Carbon with 6 protons and 6 neutrons in its nucleus (Carbon-12) was given a mass of exactly 12. It was a good choice, since it was in between the two previously used standards, and meant that nothing had to change too much.
| Which of the following statements is correct? |
- The third reason is the most important of all. If a hydrogen atom has only one proton, and carbon-12 has 6 protons and 6 neutrons to make up its mass of twelve, why isn't the mass of hydrogen 1/12 of that of carbon-12?
Mass of 1 hydrogen atom Mass of sub-atomic particles Mass of 1 carbon-12 atom 1.00794 6 protons = 6 x 1.007277 6.043662 6 neutrons = 6 x 1.008665 6.051990 6 electrons = 6 x 0.000548 0.003288 Total 12.098940 12.0 exactly If you think about it, Hydrogen at 1.00794 is more than 1/12 of the weight of carbon-12 (as you can see from the above table, if you multiply 12 times the mass of a single hydrogen atom it comes to more than 12). The reason for this effect is nuclear binding energy. After all, the protons in the nucleus are all positive, and so the nucleus should just repel itself apart. It doesn't of course, so something must be 'binding' it together. This nuclear binding energy makes the mass of all atoms (except hydrogen-1, which only has 1 proton) slightly lighter that what you'd get by adding up the mass of the sub-atomic particles. Einstein's famous equation E = mc2 shows us that we can get the necessary binding energy from the mass of the sub-atomic particles. So the mass of any multi-nucleon atom is less than the sum of the weights of its separated parts. Its this change in mass when the nucleus changes size that is the source of the enormous amount of energy in nuclear reactions.
So we could have set hydrogen to be exactly 1, but then we'd have had to really revise the atomic weight table back in 1961. If hydrogen was assigned a mass of 1 exactly, then oxygen would have become 15.87, quite a difference from the mass chemists were using. Choosing carbon-12 as the reference standard meant the least change was necessary. Still, if you do really accurate calculations based on the old and the new scale you can see some differences. For example, on the pre-1961 atomic weight scale the molecular weight of table salt, Sodium chloride NaCl would have been 58.45. On today's scale it is 58.44. The difference is just 0.02%, so for most purposes it wouldn't matter.
Hold it! You just used the term molecular weight. Isn't that wrong? Yes, of course it is, but for Sodium chloride, we shouldn't even use the term molecular mass. Instead we should use the term 'formula mass', because Sodium Chloride really isn't a molecule of NaCl.
Copyright © 1998 - 2008 David Dice