Physics 249 Homework 7
Due Nov 2 nd
1) For l=1 m=0, 1 and l=2 m=0 show that the spherical harmonic solutions satisfy the angular momentum operator with the correct eigenvalue.
𝑌
11
3
(𝜃, 𝜙) = −√
8𝜋 𝑠𝑖𝑛𝜃𝑒 𝑖𝜙
For this wave function we want to demonstrate:
𝐿 2 𝑌 𝑙𝑚
(𝜃, 𝜙) = −ℏ 2 [
1 𝑠𝑖𝑛𝜃
𝜕
𝜕𝜃
(𝑠𝑖𝑛𝜃
𝜕𝜓
) +
𝜕𝜃
1 𝑠𝑖𝑛 2 𝜃
𝜕 2
𝜕𝜙 𝜓
2
] 𝑌 𝑙𝑚
(𝜃, 𝜙) = ℏ 2 𝑙(𝑙 + 1)𝑌 𝑙𝑚
(𝜃, 𝜙)
−ℏ 2
1
[ 𝑠𝑖𝑛𝜃
𝜕
𝜕𝜃
(𝑠𝑖𝑛𝜃
𝜕𝜓
) +
𝜕𝜃
1 𝑠𝑖𝑛 2 𝜃
𝜕 2
𝜕𝜙 𝜓
2
] (−√
3
8𝜋 𝑠𝑖𝑛𝜃𝑒 𝑖𝜙 )
= ℏ 2
3
√
8𝜋
1
[ 𝑠𝑖𝑛𝜃
𝜕
𝜕𝜃
(𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃𝑒 𝑖𝜙 ) −
1 𝑠𝑖𝑛 2 𝜃 𝑠𝑖𝑛𝜃𝑒 𝑖𝜙 ]
= ℏ 2 √
3
8𝜋
1
[ 𝑠𝑖𝑛𝜃
(−𝑠𝑖𝑛 2 𝜃𝑒 𝑖𝜙 + 𝑐𝑜𝑠 2 𝑒 𝑖𝜙 ) −
1 𝑠𝑖𝑛𝜃 𝑒 𝑖𝜙 ]
= ℏ 2
3
√
8𝜋
1
[ 𝑠𝑖𝑛𝜃
(−𝑠𝑖𝑛 2 𝜃𝑒 𝑖𝜙 + 𝑒 𝑖𝜙 −𝑠𝑖𝑛 2 𝜃𝑒 𝑖𝜙 − 𝑒 𝑖𝜙 )]
= ℏ 2 √
3
8𝜋
[−2𝑠𝑖𝑛𝜃𝑒 𝑖𝜙 ] = −ℏ 2
3
2√
8𝜋 𝑠𝑖𝑛𝜃𝑒 𝑖𝜙 = ℏ 2 𝑙(𝑙 + 1)𝑌
11
(𝜃, 𝜙), 𝑙 = 1
𝑌
10
3
(𝜃, 𝜙) = √
4𝜋 𝑐𝑜𝑠𝜃
For this wave function we want to demonstrate:
𝐿 2 𝑌 𝑙𝑚
(𝜃, 𝜙) = −ℏ 2
1
[ 𝑠𝑖𝑛𝜃
𝜕
𝜕𝜃
(𝑠𝑖𝑛𝜃
𝜕𝜓
) +
𝜕𝜃
1 𝑠𝑖𝑛 2 𝜃
𝜕 2
𝜕𝜙 𝜓
2
] 𝑌 𝑙𝑚
(𝜃, 𝜙) = ℏ 2 𝑙(𝑙 + 1)𝑌 𝑙𝑚
(𝜃, 𝜙)

= −ℏ 2
1
[ 𝑠𝑖𝑛𝜃
𝜕
𝜕𝜃
(𝑠𝑖𝑛𝜃
𝜕𝜓
) +
𝜕𝜃
1 𝑠𝑖𝑛 2 𝜃
𝜕 2
𝜕𝜙 𝜓
2
] √
3
4𝜋 𝑐𝑜𝑠𝜃
= −ℏ 2 √
3
4𝜋
1
[ 𝑠𝑖𝑛𝜃
𝜕
𝜕𝜃
(𝑠𝑖𝑛𝜃
𝜕𝜓
) +
𝜕𝜃
1 𝑠𝑖𝑛 2 𝜃
𝜕 2
𝜕𝜙 𝜓
2
] 𝑐𝑜𝑠𝜃
= −ℏ 2 √
3
4𝜋
1
[ 𝑠𝑖𝑛𝜃
𝜕
𝜕𝜃
(−𝑠𝑖𝑛 2 𝜃)]
= ℏ 2
3
√
4𝜋
1
[ 𝑠𝑖𝑛𝜃
(2𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃)] = ℏ 2
3
2√
4𝜋 𝑐𝑜𝑠𝜃 = ℏ 2 𝑙(𝑙 + 1)𝑌
10
(𝜃, 𝜙), 𝑙 = 1
For this wave function we want to demonstrate:
5
𝑌
20
(𝜃, 𝜙) = √
16𝜋
(3𝑐𝑜𝑠 2 𝜃 − 1)
𝐿 2 𝑌 𝑙𝑚
(𝜃, 𝜙) = −ℏ 2
1
[ 𝑠𝑖𝑛𝜃
𝜕
𝜕𝜃
(𝑠𝑖𝑛𝜃
𝜕𝜓
) +
𝜕𝜃
1 𝑠𝑖𝑛 2 𝜃
𝜕 2
𝜕𝜙 𝜓
2
] 𝑌 𝑙𝑚
(𝜃, 𝜙) = ℏ 2 𝑙(𝑙 + 1)𝑌 𝑙𝑚
(𝜃, 𝜙)
= −ℏ 2
1
[ 𝑠𝑖𝑛𝜃
𝜕
𝜕𝜃
(𝑠𝑖𝑛𝜃
𝜕𝜓
) +
𝜕𝜃
1 𝑠𝑖𝑛 2 𝜃
𝜕 2
𝜕𝜙 𝜓
2
] √
5
16𝜋
(23𝜃 − 1)
= −ℏ 2
5
√
16𝜋
1
3 [ 𝑠𝑖𝑛𝜃
𝜕
𝜕𝜃
(−2𝑠𝑖𝑛 2 𝜃𝑐𝑜𝑠𝜃)]
= ℏ 2
5
√
16𝜋
1
6 [ 𝑠𝑖𝑛𝜃
(−𝑠𝑖𝑛 3 𝜃 + 2𝑠𝑖𝑛𝜃𝑐𝑜𝑠 2 𝜃)]
= ℏ 2 √
5
16𝜋
6[(−𝑠𝑖𝑛 2 𝜃 + 2𝑐𝑜𝑠 2 𝜃)] = ℏ 2 √
5
16𝜋
6[(3𝑐𝑜𝑠 2 𝜃 − 1)]

= ℏ 2
5
6√
16𝜋
[(3𝑐𝑜𝑠 2 𝜃 − 1)] = ℏ 2 𝑙(𝑙 + 1)𝑌
20
(𝜃, 𝜙), 𝑙 = 2
2) For the radial wave equations n=2, l=0,1 show that they are solutions of the radial wave equation with the correct energy and constant term (from the separation of variables) involving the quantum number l.
1 𝜕
𝑅(𝑟) 𝜕𝑟
(𝑟 2
𝜕
𝜕𝑟
(𝑟 2
𝜕𝑅
𝜕𝑟
𝜕𝑅
𝜕𝑟
) +
) +
2𝜇 ℏ 2 𝑟 2 (𝐸 − 𝑉(𝑟)) = 𝐶 𝑟
2𝜇 𝑟 2 (𝐸 − 𝑉(𝑟))𝑅(𝑟) = 𝑙(𝑙 + 1)𝑅(𝑟) ℏ 2
𝑅
20
(𝑟) =
1
√2𝑎 3
0
(1 − 𝑟
2𝑎
0
) 𝑒 −𝑟/2𝑎
0 just consider the radial derivative term
𝜕
𝜕𝑟
(𝑟 2
𝜕𝑅
20
𝜕𝑟
(𝑟)
)
=
𝜕
𝜕𝑟
(𝑟 2
1
[
√2𝑎 3
0
(−
1
2𝑎
0
) 𝑒 −𝑟/2𝑎
0
1
+
√2𝑎 3
0
(1 − 𝑟
2𝑎
0
) (−
1
2𝑎
0
) 𝑒 −𝑟/2𝑎
0 ])
=
𝜕
𝜕𝑟
1
([−
2𝑎
0
1
√2𝑎 3
0
(2𝑟 2 − 𝑟 3
2𝑎
0
1
= −
2𝑎
0
1
√2𝑎 3
0
(4𝑟 −
3𝑟 2
2𝑎
0
) 𝑒
− 𝑟
2𝑎
0
) 𝑒 −𝑟/2𝑎
0 ])
1
+ (
2𝑎
0
)
2
1
√2𝑎 3
0
(2𝑟 2 − 𝑟 3
2𝑎
0
) 𝑒 −𝑟/2𝑎
0
= [−
1
2𝑎
0
(4𝑟 −
3𝑟 2
2𝑎
0
1
) + (
2𝑎
0
)
2
(2𝑟 2 − 𝑟 3
2𝑎
0
)]
1
√2𝑎 3
0 𝑒
− 𝑟
2𝑎
0
2
= [− 𝑎
0
1 𝑟 + 5 (
2𝑎
0
)
2 𝑟 2
1
− (
2𝑎
0
)
2 𝑟 3
2𝑎
0
]
1
√2𝑎 3
0 𝑒
− 𝑟
2𝑎
0

= −
2 𝑎
0 𝑟 [1 −
5 1
4 2𝑎
0 𝑟 +
1
4
1
(
2𝑎
0
)
2 𝑟 2 ]
1
√2𝑎 3
0 𝑒
− 𝑟
2𝑎
0
2
= − 𝑎
0 𝑟 [(1 − 𝑟
2𝑎
0
) (1 −
1
4 𝑟
2𝑎
0
)]
1
√2𝑎 3
0 𝑒
− 𝑟
2𝑎
0
2
= (− 𝑎
0 𝑟 + 𝑟 2
4𝑎 2
0
)
1
√2𝑎 3
0
(1 − 𝑟
2𝑎
0
) 𝑒
− 𝑟
2𝑎
0
2
= (− 𝑎
0 𝑟 + 𝑟 2
4𝑎 2
0
) 𝑅
20
(𝑟) then inserting this into the full radial wave equation
2
(− 𝑎
0 𝑟 + 𝑟 2
4𝑎 2
0
) 𝑅
20
(𝑟) +
2𝜇 ℏ 2 𝑟 2 (𝐸 − 𝑉(𝑟))𝑅
20
(𝑟) = 𝑙(𝑙 + 1)𝑅
20
(𝑟)
2
(− 𝑎
0 𝑟 + 𝑟 2
4𝑎 2
0
) +
2𝜇 𝑟 2 (𝐸 − 𝑉(𝑟)) = 𝑙(𝑙 + 1) ℏ 2 for l=0
(−
2 𝑎
0 𝑟 + 𝑟 2
4𝑎 2
0
) + ℏ 2
𝐸 = −
2𝜇𝑛 2 𝑎 2
0
2𝜇 ℏ
= −
2 𝑟 2 (𝐸 − 𝑉(𝑟)) = 0 ℏ 2
8𝜇𝑎 2
0 which is correct. for n=2 inserting energy and potential
2
(− 𝑎
0 𝑟 + 𝑟 2
4𝑎 2
0
) +
2𝜇 𝑟 2 ℏ 2 ℏ 2
(−
2𝜇4𝑎 2
0
2
(− 𝑎
0
2
(− 𝑎
0 𝑟 + 𝑟 2
4𝑎 2
0
) + (− 𝑟 2
4𝑎 2
0 𝑟 + 𝑟 2
4𝑎 2
0
) + (− 𝑟 2
4𝑎 2
0
+
+
2𝜇 𝑘𝑒 2 ℏ
2
2 1
+ 𝑘𝑒 2
) = 0 𝑟) = 0 𝑎
0 𝑟) = 0 𝑟

The two sides are equal and the wave function satisfies the radial wave equation.
𝑅
21
(𝑟) =
1 𝑟
2√6𝑎 3
0 𝑎
0 𝑒 −𝑟/2𝑎
0 considering the radial derivative term
𝜕
𝜕𝑟
(𝑟 2
𝜕𝑅
21
𝜕𝑟
(𝑟)
)
1 1
=
2√6𝑎 3
0 𝑎
0
𝜕
𝜕𝑟
(𝑟 2 (1 − 𝑟
2𝑎
0
) 𝑒 −𝑟/2𝑎
0 )
=
1 1
2√6𝑎 3
0 𝑎
0
𝜕
𝜕𝑟
((𝑟 2 − 𝑟 3
2𝑎
0
) 𝑒 −𝑟/2𝑎
0 )
=
1 1
((2𝑟 −
3𝑟 2
2𝑎
0
) −
1
2𝑎
0
(𝑟 2 − 𝑟 3
2𝑎
0
)) 𝑒
2√6𝑎 3
0 𝑎
0
=
1 𝑟
2√6𝑎 3
0 𝑎
0 𝑒 −𝑟/2𝑎
0 (2 − 𝑎
2
0
1 𝑟 + (
2𝑎
0
2
= (2 − 𝑎
0
1 𝑟 + (
2𝑎
0
)
2 𝑟 2 ) 𝑅
21
(𝑟)
)
2 𝑟 2 )
−𝑟/2𝑎
0 then inserting this into the full radial wave equation
(2 − 𝑎
2
0
1 𝑟 + (
2𝑎
0
)
2 𝑟 2 ) 𝑅
21
(𝑟) +
2𝜇 ℏ 2 𝑟 2 (𝐸 − 𝑉(𝑟))𝑅
21
(𝑟) = 𝑙(𝑙 + 1)𝑅
21
(𝑟)
2 − 𝑎
2
0
1 𝑟 + (
2𝑎
0
)
2 𝑟 2 +
2𝜇 ℏ 2 𝑟 2 (𝐸 − 𝑉(𝑟)) = 𝑙(𝑙 + 1) putting in l=1 the constant terms are equal.
For the energy ℏ 2
𝐸 = −
2𝜇𝑛 2 𝑎 2
0
= − ℏ 2
8𝜇𝑎 2
0

The term, which gives the potential is the same as above as so is correct. Therefore, wave function satisfies the radial wave equation!
3) Assuming the electron to be a classical particle, a sphere of radius 10 -15 m with a uniform mass density, using the magnitude of the spin angular momentum to compute the rotational speed at the equator.
Accepting either a relativistic or non-relativistic version. Though you should use the non relativistic version since we are posing a classical (non relativistic – non quantum) source for angular momentum.
Using the moment of inertia for a uniform sphere.
𝐿 = 𝐼𝜔 =
2
5 𝑚𝑟 2 𝑣 𝑟
𝐿 → 𝑆 = ℏ√𝑠(𝑠 + 1) =
√3 ℏ
2 𝑣 =
5 1
2 𝑚𝑟
𝑆 =
5 1
2 𝑚𝑟
√3 ℏ =
2
5√3
4
1.055𝑥10 −34 𝐽𝑠
9.11𝑥10 −31 𝑘𝑔10 −15 𝑚
= 2.51𝑥10 11 𝑚/𝑠 = 837𝑐
Assume instead that the ”electron” is two equal point like masses (each with ½ the electron mass) rotating around each other bound by some unknown force and find the distance between them that give a velocity just under the speed of light. Where would you have to measure the electric field of that system to spot a 1% difference compared to the electric field of a point like charge.
This would be an answer more consistent with our understanding of fundamental particles as point like and would assume the electron had two parts.
𝐿 = 𝐼𝜔 =
1
2 𝑚𝑟 2 𝑣 𝑟
= 𝑆
1 𝑣 = 2 𝑚𝑟
1
𝑆 = 2 𝑚𝑟
√3 ℏ = √3
2
1 𝑟
1.055𝑥10 −34 𝐽𝑠
9.11𝑥10 −31 𝑘𝑔
~3.00𝑥10 8 d = 2r = 6.69x10-13m
This is only ~2 orders of magnitude less than the bohr radius so quite large.
The electric field of a point charge is given by

𝐸 = 𝑘𝑒
|𝑟⃗| 2 𝑟̂ and for two point charges
𝐸 =
1
2 𝑘𝑒
|𝑟⃗
1
| 2
1
2 𝑘𝑒
|𝑟⃗
2
| 2 𝑟 ̂
2
For the electric field or force to vary by 1% r would have to be different by 10%.
If we consider putting the charges in the x axis symmetrically around x=0 then if we look on the z axis we need 𝑧 tan(𝜃) = = 𝑧
= 0.9
𝑥 1
2 𝑑 about 3x10-13 meters up.
On the x axis you see the same magnitude change at a substantially larger distance.
On the x axis on charge will be ½d closer and the other ½d farther. Lets calculate what percentage ,p, of the distance on the x axis r has to be to cause a 1% difference
𝐸 =
1
2 𝑘𝑒
(𝑥 − 𝑝𝑥)
+
1
2 𝑘𝑒
(𝑥 + 𝑝𝑥) 2
= 0.99
𝑘𝑒 𝑥 2
1
(1 − 𝑝)
+
1
(1 + 𝑝) 2
= 0.99
p~6%, ½d = 0.06x, x = d/0.12. You can detect at ~8 times d on the x axis. About 5x10-
12 m!
We would probably expect to see the evidence of this.
4) A hydrogen atom has an electron in the l=2 states. a) What are the possible values of j and for each j mj. b) What are the values of the total angular momentum, in terms of hbar. l = 2 and s = 1/2 j = |l-s|…|l+s| = 3/2 5/2 j = 3/2, mj = -3/2 -1/2 1/2 3/2 j = 5/2, mj = -5/2 -3/2 -1/2 1/2 3/2 5/2
J = sqroot(j(j+1))hbar, j = 5/2: sqroot(35)/2 hbar, j=3/2, sqroot(15)/2 hbar

5) If the 3s electron in sodium did not penetrate the core its energy would be (-13.6/9)eV.
Explain why. Given that it’s ionization potential is 5.14 V calculate Zeff.
If it did not penetrate the core it would just see one proton worth of charge. Then it’s energy would be
𝐸 𝑛
= −
𝐸
1 𝑛 2
= −
13.6𝑒𝑉
3 2
𝐸 𝑛
= −
𝑍 2 𝑒𝑓𝑓
𝐸
1 𝑛 2
= −4.407
= −
𝑍 2 𝑒𝑓𝑓
∗ 13.6
3 2
= −5.14
𝑍 𝑒𝑓𝑓
= 1.84