Quantization of Angular Momentum

*Quantization of Angular Momentum

55 ••

55 •• [SSM][SSM] The The z z component of the spin of an electron iscomponent of the spin of an electron is

−−

¹¹₂₂^h^h, but the, but the magnitude

magnitude of of the the spin spin vector vector is is 0.750.75^h^h. What is the angle between the electron’s. What is the angle between the electron’s spin angular momentum vector and the positive

spin angular momentum vector and the positive z- z-axis?axis?

Picture the Problem

Picture the Problem The electron’sThe electron’s spin angular momentum vector is spin angular momentum vector is related to its

related to its z z component as shown component as shown in the diagram. The angle between in the diagram. The angle between

ssr

rand the +and the + z- z-axis isaxis is φ φ ..

2 2 1 1

−

− ^s^s

θ θ

z z

7 7 5 5 . 0 . 0

φ φ

Express

Express φ φ in terms of in terms of θ θ to obtain: to obtain: φ φ

==

118800

°° −−

θ θ Using trigonometry, relate the

Using trigonometry, relate the magnitude of

magnitude of ss^r^rto itsto its

−−

z z component: component:

⎟⎟⎟⎟

⎠ ⎠ ⎞

⎜⎜⎜⎜ ⎞

⎝

⎛

==

⁻⁻

⎛

h h h h

75 75 ..

0 cos 0

cos ²²

1 1 1

θ 1

Substitute for

Substitute for θ θ in the expression forin the expression for φ

φ to obtain: to obtain:

⎟⎟⎟⎟ == °°

⎠ ⎠ ⎞

⎜⎜⎜⎜ ⎞

⎝

⎛

−− ⎛

°°

==

⁻⁻ 125125

75 75 ..

0 cos 0

cos 180

180 ²²

1 1 1 1

h h h

θ h

56 ••

56 •• Show Show that that the the energy energy difference difference between between one one rotational rotational state state of of aa molecule and the next

molecule and the next higher state is proportional tohigher state is proportional to^l^l + 1. + 1.

Picture the Problem

Picture the Problem Equation 10-29Equation 10-29aa describes the quantization of rotational describes the quantization of rotational energy. We can show that the energy difference between a given state and the energy. We can show that the energy difference between a given state and the next

next higher higher state state is is proportional proportional to to ^l^l

++

11 by by using using Equation Equation 10-2710-27aa to express the to express the energy difference.

energy difference.

From Equation 10-29

From Equation 10-29aa we have: we have: K K _l_l

==

^l^l

( (

^l^l

++

11 E

))

E ₀₀_r_r Using this equation, express the

Using this equation, express the difference between one rotational difference between one rotational state and the next higher state:

state and the next higher state:

(

( ))( ( ) ) ( ( )) ( ( ))

₀₀_r_r

r r 0 0 r

r 0 0

1 1 2 2

1 1 2

2 1

1 E E

E E E

E E

++

==

++

−−

++

==

Δ Δ

l l

l l l l l

l l

57 ••

57 •• [SSM][SSM] You You work work in in a a bio-chemical bio-chemical research research lab, lab, where where you you areare investigating the rotational energy levels of the

investigating the rotational energy levels of the HBr molecule. After consultingHBr molecule. After consulting the periodic chart, you

the periodic chart, you know that the mass of the bromine aknow that the mass of the bromine atom is 80 times that oftom is 80 times that of the hydrogen atom. Consequently, in calculating the rotational motion of the the hydrogen atom. Consequently, in calculating the rotational motion of the molecule, you assume, to a good approximation, that the Br nucleus remains molecule, you assume, to a good approximation, that the Br nucleus remains stationary as the H atom (mass 1.67

stationary as the H atom (mass 1.67

××

10 10^–27^–27 kg) revolves around it. You a kg) revolves around it. You also knowlso know

that the separation between the H atom and bromine nucleus is 0.144 nm.

Calculate (

Calculate (aa) the moment of inertia of the HBr molecule about the bromine) the moment of inertia of the HBr molecule about the bromine nucleus, and (

nucleus, and (bb) the rotational energies for the bromine nucleus’s) the rotational energies for the bromine nucleus’s ground state ground state (lowest energy)

(lowest energy) ^l^l = 0, and the next two states of higher energy (called the first and = 0, and the next two states of higher energy (called the first and second

second excited statesexcited states) described by) described by ^l^l = 1, and = 1, and^l^l = 2. = 2.

Picture the Problem

Picture the Problem The rotational energies of HBr molecule are related toThe rotational energies of HBr molecule are related to ^l^l and

and E E according to₀₀_r_raccording to K K _l_l

==

^l^l

( (

^l^l

++

11 E

))

E ₀₀_r_r w whheerree E E ₀₀_r_r

==

^h^h²² 22 I I ..

((aa) Neglecting the motion of the) Neglecting the motion of the bromine molecule:

bromine molecule:

2 2 H H 2 2 p p HBr

HBr mm r r mm r r I

≈≈ ==

Substitute numerical values and Substitute numerical values and evaluate

evaluate I I HBr HBr ::

( ( ))( ( ))

2 2 47

2 9 2 9 27

27 HBr

HBr

m m kg kg 10 10 3.46 3.46

m m kg kg 10 10 3.463 3.463

m m 10 10 0.144 0.144 kg

kg 10 10 1.67 1.67

⋅⋅

××

==

⋅⋅

××

==

××

≈≈

−−

I −−

((bb) Relate the rotational energies to) Relate the rotational energies to

landand E E ::₀₀_r_r ^K^Kll

⁼⁼

^l^l

( (

^l^l

⁺⁺

¹^1 E

))

^E₀₀_r_rwherewhere

HBr HBr

2 2 r

r 0

0 22 I I E

==

^h^h

Substitute numerical values and Substitute numerical values and evaluate

evaluate E E ::₀₀_r_r

( ( ))

meV meV 003 003 ..

1 1

JJ 10 10 1.602 1.602

eV eV 1 JJ 1

10 10 1.607 1.607

m m kg kg 10 10 3.463 3.463 2

ss JJ 10 10 1.055 1.055 2

19 19 22

2 2 47

2 34 2 34 2

2 r r 0 0

==

×× ××

××

==

⋅⋅

××

⋅⋅

== ××

==

−−

I E I

E ^h^h

Evaluate

Evaluate E E 00 to obtain: to obtain: E E ₀₀

==

K K ₀₀

==

11..0000meVmeV

Evaluate

Evaluate E E 11 to obtain: to obtain:

( ( ))( ( ))

meV meV 01 01 ..

2 2

meV meV 003 003 ..

1 1 1 1 1

1 1

1 1 1

==

++

==

K K E E

Evaluate

Evaluate E E 22 to obtain: to obtain:

( ( ))( ( ))

meV meV 02 02 ..

6 6

meV meV 003 003 ..

1 1 1 1 2 2 2

2 2

2 2 2

==

++

==

K K E E

58 •••

58 ••• The equilibrium separation between the nuclei of the nitrogenThe equilibrium separation between the nuclei of the nitrogen molecule (N

molecule (N₂₂) is 0.110 nm and the mass of each nitrogen nucleus is 14.0 u, where) is 0.110 nm and the mass of each nitrogen nucleus is 14.0 u, where u = 1.66

u = 1.66

××

10 10^–27^–27 kg. For rotational energies, the total energy is due to rotational kg. For rotational energies, the total energy is due to rotational kinetic energy. (

kinetic energy. (aa) Approximate the nitrogen molecule as a ) Approximate the nitrogen molecule as a rigid dumbbell of tworigid dumbbell of two equal point masses and calculate

equal point masses and calculate the moment of inertia about its center of the moment of inertia about its center of mass.mass.

((bb) Find the energy) Find the energy E E ll of the lowest three energy levels using of the lowest three energy levels using called a

called a photon photon when they make a when they make a transitiontransition from a higher energy state to a lower from a higher energy state to a lower one. Determine the energy of a photon emitted when a nitrogen molecule drops one. Determine the energy of a photon emitted when a nitrogen molecule drops from the

from the^l^l = 2 to the = 2 to the^l^l = 1 state. Visible light photons each have between 2 and = 1 state. Visible light photons each have between 2 and 3 eV of energy. Are these photons in the visible region?

3 eV of energy. Are these photons in the visible region?

Picture the Problem

Picture the Problem We can use the definition of the moment of inertia of pointWe can use the definition of the moment of inertia of point particles to calculate the

particles to calculate the rotational inertia of the rotational inertia of the nitrogen molecule. The nitrogen molecule. The rotationalrotational energies of nitrogen molecule are related to

((aa) Using a rigid dumbbell model,) Using a rigid dumbbell model, express and evaluate the moment of express and evaluate the moment of inertia of the nitrogen molecule inertia of the nitrogen molecule about its center of mass:

about its center of mass:

Substitute numerical values and Substitute numerical values and evaluate

((bb) Relate the rotational energies) Relate the rotational energies to

Substitute numerical values and Substitute numerical values and evaluate

Evaluate

Evaluate E E 22 to obtain: to obtain:

( ( ))( ( ))

meV meV 48 48 ..

1 1

meV meV 0.2474 0.2474 1

1 2 2 2

2 2

==

++

==

E E

((cc) The energy of a photon emitted) The energy of a photon emitted when a nitrogen molecule drops when a nitrogen molecule drops from the

from the^l^l = 2 to the = 2 to the^l^l = 1 state is: = 1 state is:

meV meV 99 99 ..

0 0

meV meV 495 495 ..

0 0 meV meV 48 48 ..

1 1

Δ 22 11 22 11

==

−−

==

−−

==

→

== E E E E

E E _l_l _l_l

No. This energy is too low to produce radiation in the visible portion of the No. This energy is too low to produce radiation in the visible portion of the

spectrum.

59 •••••• Consider aConsider a transition transition from a lower energy state to a higher one. That from a lower energy state to a higher one. That is, the absorption of a qua

is, the absorption of a quantum of energy resulting in an increase ntum of energy resulting in an increase in the rotationalin the rotational energy of an N

energy of an N22 molecule (see Problem 64). Suppose such a molecule, initially in molecule (see Problem 64). Suppose such a molecule, initially in its ground rotational state, was exposed to photons each with energy equal to the its ground rotational state, was exposed to photons each with energy equal to the three times the energy of its first excited state.

three times the energy of its first excited state. ((aa) Would the molecule be able to) Would the molecule be able to absorb this photon energy? Explain why or why not and if it can, determine the absorb this photon energy? Explain why or why not and if it can, determine the energy level to which it goes. (

energy level to which it goes. (bb) To make a transition from its ground state to its) To make a transition from its ground state to its second excited state requires how

second excited state requires how many times the energy of the many times the energy of the first excited state?first excited state?

Picture the Problem

Picture the ProblemThe rotational energies of a nitrogen molecule depend onThe rotational energies of a nitrogen molecule depend on the quantum number

the quantum number^l^l according to according to E E _l_l

==

L L²² //22 I I

==

^l^l

( (

^l^l

++

))

^h^h²² //22 I I ..

((aa) No. None of the allowed values of) No. None of the allowed values of E E are equal tollare equal to E E 0r 0r ..

((bb) The upward transition from the) The upward transition from the ground state to the second excited ground state to the second excited state requires energy given by:

state requires energy given by:

0 0 2 2 2 2 0

Δ 0

Δ E E _l_l₌₌ _→_→_l_l₌₌

==

E E

−−

E E

Set this energy difference equal to a Set this energy difference equal to a constant

constant nn times the energy of the 1times the energy of the 1^st^st excited state:

excited state:

1 1 0

0 2

2 E E nE nE E

− − == ⇒ ⇒

1 1

0 0 2 2

E E

E E E n E

−−

==

Substitute numerical values and Substitute numerical values and evaluate

evaluate nn::

( ( ))

( (

¹¹ ¹¹

))

²²^..⁵⁵

1 1 2 2 2 2

r r 0 0

r r 0 0 r r 0

==

++

−−

== ++

E E

E E E n E

In document Cap10 (Page 43-46)