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Question:
NOTE: First printing,2 in the denominator of the second equation should be 4.
An electron in an atom will experience an internal magnetic field Bint because, from the
electron’s reference frame, it is the positive nucleus that is orbiting the electron. The electron
will “see” the nucleus, take as charge +e, circling around it, which is equivalent to a current I
= +ef where f is the electron’s frequency of rotation around the nucleus. The current I
generates the internal magnetic field Bint at the electron. From electromagnetism texts, Bint
is given by
where r is the radius of the electron’s orbit and µo is the absolute permeability. Show that
Consider the hydrogen atom with Z = 1, 2p orbital, n = 2, l = 1, and take r≈ n2ao. Calculate Bint.
The electron’s spin magnetic moment μspin will couple with this internal field, which
means that the electron will now possess a magnetic potential energy ESL that is due to the
coupling of the spin with the orbital motion, called spin-orbit coupling. ESL will be either
negative or positive, with only two values, depending on whether the electron’s spin
magnetic moment is along or opposite Bint, Take z along Bint so that ESL = ─Bintμspin, z
where µspin, z is µspin along z, and then show that the energy E2 of the 2p orbital splits into
two closely separated levels whose separation is
Calculate ΔESL in eV and compare it with E2(n = 2) and the separation ΔE = E2 - E1. (The
exact calculation of ESL is much more complicated, but the calculated value here is
sufficiently close to be useful.) What is the effect of ESL on the observed emission spectrum
from the H-atom transition from 2p to 1s? What is the separation of the two wavelengths?
The observation is called fine structure splitting.
Answer:
Step 1 of 10
Consider the expression for the orbital angular momentum.
…… (1)
Here,
is the mass of electron in free space,
is the angular frequency, and
the orbit (current loop).
Consider the expression for the angular frequency.
…… (2)
Here,
is the frequency.
Substitute
for
in equation (1).
…… (3)
Rearrange the equation (3) to find the frequency f.
…… (4)
Consider the expression for internal magnetic field.
…… (5)
Here,
I is the current and
Substitute
is the absolute permeability.
for I in equation (5).
…… (6)
Step 2 of 10
Substitute equation (4) in equation (6).
…… (7)
Substitute
for r in equation (7).
…… (8)
Consider the expression for the angular momentum L.
is the radius of
…… (9)
Here,
is the Planck’s constant divided by
and
is the wavelength.
Step 3 of 10
Substitute equation (9) in (8).
…… (10)
Substitute 2 for n, 1 for
for
,
,
for
for
for e,
,
, and
for
…… (11)
Substitute the unit
for C and
Thus, the internal magnetic field is
for J in equation (11).
.
Step 4 of 10
Consider the expression for the spin magnetic moment.
…… (12)
Consider the expression for the spin magnetic moment along the z-axis.
…… (13)
Here,
is the electron’s spin.
Substitute equation (13) in equation (12).
in equation (10).
…… (14)
Consider the expression for the magnetic potential energy.
…… (15)
Substitute equation (14) in equation (15).
Step 5 of 10
Since, the magnetic potential energy has only two values corresponding to
and
. Therefore, the difference between the two values is,
…… (16)
for e,
Substitute
for
for
,
for
, and
in equation (16).
…… (17)
Substitute the unit
for Wb,
for J, and
for C in equation (17).
…… (18)
Substitute
for1 J in equation (18) to find the magnetic potential energy
in eV.
Thus, the magnetic potential energy
Step 6 of 10
is
.
Consider the expression for the ionization energy for the nth shell.
…… (19)
Here,
is the atomic number of atom.
Consider for hydrogen atom.
Substitute 2 for n and in equation (19).
Substitute 1 for n and in equation (19).
Step 7 of 10
Therefore, the change in ionization energy is,
…… (20)
Substitute
for
and
Thus, the magnetic potential energy
for
in equation (20).
is much less than
or
.
Figure 1 shows the energy levels of hydrogen atom with spin orbital coupling.
Step 8 of 10
In Figure 1, the transmission must be two wavelength separated by the magnetic potential
energy
for the ionization energy of the 2nd shell. Therefore,
and
two levels namely
. That is, the ionization energy of
can be separated into
and
are
for
, and
and
Consider the expression for the emission wavelength
.
…… (21)
Substitute
for
in equation (21).
…… (22)
for c,
Substitute
for
for
,
in equation (22).
…… (23)
Substitute
for1 J in equation (23).
Step 9 of 10
Similarly, consider the expression for the emission wavelength
…… (24)
Substitute
for
in equation (24).
.
…… (25)
for c,
Substitute
for
for
,
for
, and
in equation (25).
…… (26)
Substitute
for1 J in equation (26).
Step 10 of 10
Consider the expression for the difference in wavelength.
…… (27)
Substitute 121.5499 nm for
and 121.5485 nm for
in equation (27).
That is, the difference in wavelength is very small.
Thus, the separation of two wavelength are
respectively.
and
,