Physical constants
Boltzmann's constant k= 1.38066E-23 J/K
k= 0.00008617 eV/K
Atomic mass unit u= 1.66E-27 kg
umev= 931.5 MeV/c^2
Wien constant 2.90E-03
Planck's constant h= 4.14E-15 eV s
h= 6.63E-34 J s
h/ 2 pi hbar 1.05E-34 J s
h/ 2 pi hbev 6.59E-16 eV s
Speed of light c= 3.00E+08 m/s
hc= 1.24E+03 eV nm
hbarc= 1.97E+02 eV nm
Electron charge e= 1.60E-19 C
electron volt eV= 1.60E-19 joule
electron mass mel 9.11E-31 kg
mel 5.11E-01 MeV/c^2
proton mass mp 1.67E-27 kg
mpev 9.38E+02 MeV/c^2
neutron mass mn 1.67E-27 kg
mnev 9.40E+02 MeV/c^2
Stefan's constant= sigma= 5.67E-08 watt/m^2K^4
Nuclear distance unit fermi 1.00E-15 m (a femtometer)
cross section unit barn= 1.00E-28 m^2
barn= 100 fm^2
common combo for scattering ke^2 1.44 MeV fm

物理常数 Boltzmann 常数 k= 1.38066E-23 J/K k= 0.00008617 eV/K 原子质量单位 u= 1.66E-27 kg umev= 931.5 MeV/c^2 Wien 常数 2.90E-03 Planck 常数 h= 4.14E-15 eV s h= 6.63E-34 J s h/ 2 pi hbar 1.05E-34 J s h/ 2 pi hbev 6.59E-16 eV s 速率 of light c= 3.00E+08 m/s hc= 1.24E+03 eV nm hbarc= 1.97E+02 eV nm 电子电荷 e= 1.60E-19 C 电子伏特 eV= 1.60E-19 joule 电子质量 mel 9.11E-31 kg mel 5.11E-01 MeV/c^2 质子质量 mp 1.67E-27 kg mpev 9.38E+02 MeV/c^2 中子质量 mn 1.67E-27 kg mnev 9.40E+02 MeV/c^2 Stefan 常数= sigma= 5.67E-08 watt/m^2K^4 核距离单位 fermi 1.00E-15 m (一个费米) 截面单位 barn= 1.00E-28 m^2 barn= 100 fm^2 常见的散射组合 ke^2 1.44 MeV fm

First Bohr orbit radius a0 a0 0.0529 nm
Bohr magnetion mub 5.79E-05 eV/T
muba 9.27E-24 A m^2 A=ampere

第一玻尔轨道半径 a0 为 0.0529 nm，玻尔磁子 mub 为 5.79×10⁻⁵ eV/T，muba 为 9.27×10⁻²⁴ A m²，其中 A=安培。

#2 a.Use radial probability to estimate energy of ground state.
Magnitude of average kinetic energy is equal to that of average potential energy.
Average PE equal to potential energy at the expectation value for radius, ke^2/<r>
The expectation value for the radius is on the order of 0.1nm
Average KE = 1.44 eV nm/0.1 nm = 14.4 eV

#2 a. 使用径向概率估计基态的能量。平均动能的大小等于平均势能的大小。平均势能等于在半径期望值处的势能，即 ke²/<r>。半径的期望值约为0.1纳米。平均动能=1.44 eV·nm/0.1 nm=14.4 eV

2.b. From the probability curve, the distribution is mostly contained within +/- a Bohr radius, so uncertainty in
the neighborhood of 1 Bohr radius.

从概率分布曲线来看，分布主要集中在±一个玻尔半径的范围内，因此不确定性约为1个玻尔半径。

2c. The uncertainty in momentum is on the order of the momentum = sqrt(2*m*<E>) = 3836.248167 eV/c
Checking for consistency with the uncertainty principle
delta x * delta p = 3836 eV/c *.0529 nm = 202.9244 eV nm/c 1.030073096 hbar
so these estimates are consistent with the uncertainty principle.

动量的不确定性约为动量 = sqrt(2*m*<E>) = 3836.248167 eV/c。检查与不确定性原理的一致性：delta x * delta p = 3836 eV/c * 0.0529 nm = 202.9244 eV nm/c = 1.030073096 hbar，这些估计与不确定性原理一致。

#3 Probability that electron in hydrogen is closer than 1 Bohr radius.
The value of the probability for any range of r is given by
P = 4[exp(-2*r/a0)*(-a0*r^2/2 - a0^2*r/2 -a0^3/4)]/a0^3 evaluated between the radius limits.
For limits 0 to a0, the terms in parentheses above reduce to -5/4 at a0 and
to -1/4 at r=0. So the probability is exp(-2)*(-5) +1 = 0.323323584

#3 电子在氢原子中处于比1个波尔半径更近的概率。对于任意半径范围r的概率值，由P = 4[exp(-2*r/a0)*(-a0*r^2/2 - a0^2*r/2 -a0^3/4)]/a0^3在半径极限之间计算得出。当极限为0到a0时，括号中的项在a0处简化为-5/4，在r=0处简化为-1/4。因此，概率为exp(-2)*(-5/4) + 1 = 0.323323584。

#4 Probability that electron is between 1 and 4 Bohr radii if it is in ground state.
The probability stated in #3 evaluated between the limits a0 and 4*a0 gives
=4*(exp(-8)*(-16/2-4/2 -1/4) -exp(-2)*(-5))= 0.662922448

#4 电子处于基态时，其位于1到4个波尔半径之间的概率。在#3中给出的概率在a0到4*a0的极限下计算得到：4*(exp(-8)*(-16/2-4/2 -1/4) -exp(-2)*(-5))= 0.662922448。

#9 Probability that 2s and 2p electrons will be found inside the nucleus.
Since the nuclear radius is 3 orders of magnitude less that the Bohr radius
the exponential terms in the wavefunctions may be set equal to 1
The 2s wavefunction integral can then be approximated by the integral of
r^2*(2 - r/a0)^2/(2*a0^3) or r^2*(1 + r^2/(4*a0^2) - r/a0)/(2*a0^3)
The integral between zero and the nuclear radius R is
P = R^3/(6*a0^3) + R^5/(40*a0^5) - R^4/(8*a0^4)
Taking the radius R=1 fermi, this becomes
P = (1E-15 m)^3/(6*(.529E-10 m))^3 = 1.12585E-15
This compares to about 1E-14 for the 1s state, so a 1s electron is 10x more likely to
be found inside the nucleus.
b. For the 2p state, the integrand with the exponential set equal to 1 is
r^4/(24*a0^5)
The integral between zero and the nuclear radius R is
R^5/(120*a0^5) = P = (1E-15 m)^5/(120*(.529E-10 m)^5 = 2.01159E-26
This is 12 orders of magnitude less than the probability for the 1s electron.

2s和2p电子处于核内的概率。由于核半径比玻尔半径小三个数量级，波函数中的指数项可以近似设为1。因此，2s波函数的积分可以近似为r²*(2 - r/a0)²/(2*a0³) 或 r²*(1 + r²/(4*a0²) - r/a0)/(2*a0³)。在零到核半径R之间的积分得到P = R³/(6*a0³) + R⁵/(40*a0⁵) - R⁴/(8*a0⁴)。取半径R=1费米，这变为P = (1E-15 m)^3/(6*(.529E-10 m))^3 = 1.12585E-15。这与1s态的约1E-14相比，说明1s电子处于核内的概率是2s电子的10倍。b. 对于2p态，将指数项设为1后的积分项为r⁴/(24*a0⁵)。在零到核半径R之间的积分得到P = R⁵/(120*a0⁵) = (1E-15 m)^5/(120*(.529E-10 m)^5) = 2.01159E-26。这比1s电子的概率小12个数量级。

#11 Values for L and Lz for 3p and 3d states.
a. 3p state
l=1, L = sqrt(l*(l+1))*hbar = sqrt(3)*hbar
ml=-1,0,1 and Lz = ml*hbar
b. 3d state
l=2, L = sqrt(l*(l+1))*hbar = sqrt(6)*hbar
ml=-2,-1,0,1,2 and Lz = ml*hbar

#11 3p 和 3d 态的 L 和 Lz 值。a. 3p 态 l=1, L = sqrt(l*(l+1))*hbar = sqrt(3)*hbar, ml=-1,0,1 和 Lz = ml*hbar；b. 3d 态 l=2, L = sqrt(l*(l+1))*hbar = sqrt(6)*hbar, ml=-2,-1,0,1,2 和 Lz = ml*hbar

#12 Quantum numbers for n=5 states
l ml
0 0
1   -1,0,1
2   -2,-1,0,1,2
3   -3,-2,-1,0,1,2,3
4   -4,-3,-2,-1,0,1,2,3,4

#12 n=5时的量子数l和ml：0 0 1 -1,0,1 2 -2,-1,0,1,2 3 -3,-2,-1,0,1,2,3 4 -4,-3,-2,-1,0,1,2,3,4

#18 Beam splitting in Stern-Gerlach experiment
a. The force on the atoms is F = -muz * dB/dz
The z component of the magnetic moment is muz = -e*hbar/(2*m) = -1 Bohr magneton
The speed of the atoms is v = sqrt(2*KE/M), where M is the mass of the silver atom.
The time that the atoms experience the force is t = L/v = L/sqrt(2*KE/M)
The lateral displacement is given by x = .5*a*t^2 = mub*dB/dz*M*L^2/(2*M*2*KE)
The separation is twice this displacement = (mub*L^2/(2*KE))*dB/dz where mub=Bohr magneton
b. Calculation of field gradient necessary to give a mm of separation of the beams on the film
The KE=2kT for particles escaping through a hole from a hot oven (not 3kT/2 !).
The field gradient is given by dB/dz = 2*KE*d/(mub*L^2) = 4kTd/(mub*L^2)
dB/dz= (4*k*1000*.001)/(mub*.05^2) = 2.38E+03 T/m

#18 施特恩-格拉赫实验中的束流分裂 a. 原子所受的力为 F = -muz * dB/dz，其中 z 方向的磁矩分量为 muz = -e*hbar/(2*m) = -1 布拉格磁矩。原子的速率为 v = sqrt(2*KE/M)，其中 M 是银原子的质量。原子经历该力的时间为 t = L/v = L/sqrt(2*KE/M)。横向位移由 x = 0.5*a*t^2 给出，即 x = mub*dB/dz*M*L^2/(2*M*2*KE)。分离距离是此位移的两倍，即分离距离为 (mub*L^2/(2*KE))*dB/dz，其中 mub=布拉格磁矩。b. 计算使束流在胶片上分离的场梯度。KE=2kT 用于粒子从热炉孔中逃逸（不是 3kT/2！）。场梯度由 dB/dz = 2*KE*d/(mub*L^2) = 4kTd/(mub*L^2) 给出，即 dB/dz= (4*k*1000*.001)/(mub*.05^2) = 2.38E+03 T/m

#21 Quantum numbers related to total angular momentum
a. The possible values of mj are -j, -j+1, … j which give a total of 2j + 1 values
b. Quantum numbers of the n=2 states in terms of hydrogen in terms of n, l, ml, and ms
(2,0,0,1/2),(2,0,0,-1/2)
(2,1,-1,1/2),(2,1,-1,-1/2),(2,1,0,1/2),(2,1,0,-1/2),(2,1,1,1/2),(2,1,1,-1/2),
Quantum numbers of the n=2 states in terms of n,l,j,mj
(2,0,1/2,1/2),(2,0,1/2,-1/2)
(2,1,3/2,3/2),(2,1,3/2,1/2),(2,1,3/2,-1/2),(2,1,3/2,-1/2),(2,1,1/2,1/2),(2,1,1/2,-1/2),
c. If mj=3/2, what are the possible values of l, ml, and ms?
This is the stretched case where l=1 and ml=1 and ms=1/2
d. If mj=1/2, what are the possible values of l, ml, and ms?
There are several ways to get mj=1/2
(l=0,ml=0,ms=1/2),(l=1,ml=1,ms=-1/2),(l=1,ml=0,ms=1/2)

#21 与总角动量相关的量子数 a. mj的可能值为 -j, -j+1, …, j，共给出2j + 1个值 b. 以氢原子的n=2状态为例，用n, l, ml, ms表示的量子数为 (2,0,0,1/2),(2,0,0,-1/2) (2,1,-1,1/2),(2,1,-1,-1/2),(2,1,0,1/2),(2,1,0,-1/2),(2,1,1,1/2),(2,1,1,-1/2)，用n,l,j,mj表示的量子数为 (2,0,1/2,1/2),(2,0,1/2,-1/2) (2,1,3/2,3/2),(2,1,3/2,1/2),(2,1,3/2,-1/2),(2,1,3/2,-1/2),(2,1,1/2,1/2),(2,1,1/2,-1/2) c. 如果mj=3/2，l, ml, ms的可能值是什么？这是伸展情况，其中l=1，ml=1，ms=1/2 d. 如果mj=1/2，l, ml, ms的可能值是什么？有几种方法可以得到mj=1/2 (l=0,ml=0,ms=1/2),(l=1,ml=1,ms=-1/2),(l=1,ml=0,ms=1/2)

#27 Zeeman splitting for hydrogen 3->2 transition
The photon energy = 13.6 eV*(1/4 - 1/9) = 1.888888889 eV
You get 7 d states and 5 p states
The width of the set of d states is then 6*mub*B and the width of the p states is 4*mub*B
The rms width of the transition is then sqrt((delta Ed)^2 + (delta Ep)^2)= sqrt(36+16)*mub*B
delta E/E = 1E-4 = sqrt(36+16)*mub*B/1.89
This corresponds to a field B = 4.53E-01 Tesla

#27 氢原子3→2跃迁的塞曼分裂。光子能量=13.6 eV*(1/4 - 1/9)=1.888888889 eV。你得到7个d态和5个p态。d态的宽度是6μ_B*B，p态的宽度是4μ_B*B。跃迁的均方根宽度是sqrt((ΔE_d)^2 + (ΔE_p)^2)=sqrt(36+16)*μ_B*B。ΔE/E=1E-4=sqrt(36+16)*μ_B*B/1.89。这对应于一个磁场B=0.453特斯拉。

#29 Modeling electron spin with a spinning sphere
Angular momentum L = I*omega = hbar/2
For a sphere, I = 2*M*r^2/5
omega = L/I = (hbar/2)/(2*M*r^2/5) = 1.45E+32 1/s
Not a good model since electron size (spatial extent, wavelength, uncertainty in position)
depends upon its momentum

#29 用旋转球体建模电子自旋角动量 L = I*omega = hbar/2 对于球体，I = 2*M*r^2/5 omega = L/I = (hbar/2)/(2*M*r^2/5) = 1.45E+32 1/s 不是一个好的模型，因为电子的尺寸（空间范围，波长，位置的不确定性）取决于其自旋

#31 2p state of hydrogen
a. most probable radius for 2p electron.
Radial part of probability = 4*pi*r^2*wavefunction^2
Take derivative to find most probable radius
d/dr(r^2*(r/a0)^2*exp(-r/a0)) = (4r^3/a0^2)*exp(-r/a0) - (r^4/a0^3)*exp(-r/a0) =0
This gives a most probable r = 4*a0

#31 氢原子的2p态 a. 2p电子的最可能半径。概率径向部分 = 4πr²×波函数² 对r求导以找到最可能的半径 d/dr(r²×(r/a0)²×exp(-r/a0)) = (4r³/a0²)×exp(-r/a0) - (r⁴/a0³)×exp(-r/a0) = 0 这给出最可能的r = 4a0

b. Probability of being within 10% of this most probable value
Integrate the probability distribution between 3.6 and 4.4 a0.
Can approximate by taking peak probability and multiplying by the interval 0.8*a0
This shakes down to (1/(24*a0^3))*(0.8*a0)*(4*a0)^4/a0^2)*exp(-4) = 0.156293452

b. 位于这个最可能值10%范围内的概率将概率分布从3.6到4.4 a0积分。可以近似为取峰值概率并乘以区间0.8*a0。这计算结果为(1/(24*a0^3))*(0.8*a0)*(4*a0)^4/a0^2)*exp(-4) = 0.156293452