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
hc hc 1.24E+03 eV nm
Speed of light c= 3.00E+08 m/s
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
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
hbar*c 3.16138E-26 J m
hbar*c 197.5863737 eV nm
h*c 1.98635E-25 J m
h*c 1241.4718 eV nm
#1 Wavefunction for a 100 eV free electron
wavefunction = A*exp (i(px - Et)/hbar)
E = 100 eV, p=sqrt(2 m E) , approx 10 keV/c
p= 10109.40156 eV/c

物理常数玻尔兹曼常数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/c² 威恩常数 2.90E-03 平克常数h= 4.14E-15 eV·s h= 6.63E-34 J·s h/2π hbar 1.05E-34 J·s h/2π hbev 6.59E-16 eV·s hc hc 1.24E+03 eV·nm 速率光c= 3.00E+08 m/s 电子电荷e= 1.60E-19 C 电子伏特eV= 1.60E-19 焦耳电子质量mel 9.11E-31 kg mel 5.11E-01 MeV 质子质量mp 1.67E-27 kg mpev 9.38E+02 MeV/c² 中子质量mn 1.67E-27 kg mnev 9.40E+02 MeV/c² 斯特藩常数= sigma= 5.67E-08 瓦特/平方米·开尔文⁴ 核距离单位费米 1.00E-15 米（一个费米）截面单位巴恩= 1.00E-28 平方米巴恩= 100 平方费米常用散射组合ke² 1.44 MeV·fm hbar·c 3.16138E-26 J·m hbar·c 197.5863737 eV·nm h·c 1.98635E-25 J·m h·c 1241.4718 eV·nm #1 100 eV自由电子波函数= A*exp (i(px - Et)/hbar) E = 100 eV, p=sqrt(2 m E) , approx 10 keV/c p= 10109.40156 eV/c

#6. Proton in 2 fermi one dimensional box, first two energies
E = (nh)^2/(8mL^2) = (nhc)^2/(8 mc^2L^2) = 5.12E+01 MeV for n=1
2.05E+02 MeV for n=2

#6. 一个质子在2费米的一维箱中，前两个能量为 E = (nh)^2/(8mL^2) = (nhc)^2/(8 mc^2L^2) = 51.2 MeV 对于n=1，205 MeV 对于n=2

#8. Electron in box, 10 eV difference between ground and first excited, find L
Since the energy is proportional to n^2, the difference between the ground and first excited
states gives 3(hc)^2/(8 mc^2L^2) = 10 eV
L = sqrt(3)*hc/sqrt(80*mc^2) 0.335913044 nm

#8. 箱中电子，基态与第一激发态之间的能量差为10 eV，求L。由于能量与n²成正比，基态与第一激发态之间的能量差为3(hc)²/(8 mc²L²) = 10 eV，解得L = sqrt(3)*hc/sqrt(80*mc²) = 0.335913044 nm

#13. Finite square well of height 10 eV. If ground energy is 1 eV, what is width of well?
The equation for the finite square well is trancendental if you are solving for energy, but
in this case the energy is given, so you can solve for the corresponding well width L.
tan(sqrt(2*m*E)*L/(2*hbar)) = tan(sqrt(2*mc^2*E)*L/(2*hbar*c))=sqrt((V0-E)/E)
tan(sqrt(2*.511E6*1)*L/(2*197)) = sqrt((10-1)/1) = 3
sqrt(2*.511E6*1)*L/(2*197) = 1.249045772
L = 0.486798384 nm

一个高度为10 eV的有限方形势阱。如果底能是1 eV，求势阱的宽度。有限方形势阱的方程在求解能量时是超越方程，但在此情况下，能量已知，因此可以求出对应的势阱宽度L。tan(sqrt(2*m*E)*L/(2*hbar)) = tan(sqrt(2*mc^2*E)*L/(2*hbar*c))=sqrt((V0-E)/E) tan(sqrt(2*0.511e6*1)*L/(2*197)) = sqrt((10-1)/1) = 3 sqrt(2*0.511e6*1)*L/(2*197) = 1.249045772 L = 0.486798384 纳米

#15. Alpha particle of energy 4 MeV, Estimate lifetime.
From Fig 7-17, p208 of Rohlf, read halflife of about 1E18 seconds

#15 能量为4 MeV的α粒子，估计其寿命。参考Rohlf第7章第17图，第208页，寿命约为1×10¹⁸秒。

#17. Probability for 3 eV electron penetrating 4 eV barrier of height 4 eV and width 0.1 nm.
The probability for penetration is of t he form exp(-2 beta x) where x is the thickness of the barrier.
The term beta is given by beta=sqrt(2 mc^2 (V0 - E))/hbar*c
beta = 5.131675921 nm^-1
The corresponding lifetime is exp(-2*5.13 nm^-1*0.1nm) = 0.358317726 seconds.

3 eV电子穿透4 eV势垒的概率形式为exp(-2 beta x)，其中x是势垒的厚度。beta项由beta=sqrt(2 mc^2 (V0 - E))/hbar*c给出，beta = 5.131675921 nm^-1。相应的寿命为exp(-2*5.13 nm^-1*0.1nm) = 0.358317726秒。

#24. Uncertainty principle and the quantum harmonic oscillator
a. delta x = sqrt(hbar/ m*omega)
b. delta p = sqrt(2*m*E) = sqrt(2*m*(hbar*omega/2)) = sqrt(m*hbar*omega)
c. For an electron where hbar*omega = 10 eV
delta x = sqrt(hbar^2*c^2/(mc^2 hbar*omega))= hbar*c/sqrt(.511E6 eV*10 eV)
delta x = 0.087147669 nm
delta p= sqrt(m*hbar*omega)=sqrt(mc^2* hbar*omega)/c =sqrt(.511E6 eV*10 eV)/c
delta p= 2260.530911 eV/c

#24. 不确定性原理和量子谐振子 a. Δx = sqrt(ħ/(mω)) b. Δp = sqrt(2mE) = sqrt(2m(hbarω/2)) = sqrt(mhbarω) c. 对于电子，其中 ħω = 10 eV Δx = sqrt(ħ²c²/(mc² ħω))= ħc/sqrt(0.511E6 eV*10 eV) Δx = 0.087147669 nm Δp= sqrt(mhbarω)=sqrt(mc²* hbarω)/c =sqrt(0.511E6 eV*10 eV)/c Δp= 2260.530911 eV/c

#26. Energy for electron in atomic sized 3-D box.
Let the wavelength of the electron equal twice the dimension of the box so that
wavelength = 2*V^(1/3) = 2*(1E-29)^(1/3)=   4.30887E-10 m
minimum momentum = sqrt(3(h/(2*V^(1/3))^2) = sqrt(3)*hc/(2*c*V^(1/3))
=sqrt(3)*1240 eV nm/(2*c*(10^-2 nm^2)^(1/3)) =    4984.469966 eV/c
b. The ground state energy is E=3*(hc)^2/(8*mc^2*V^(2/3)) =
=3*(1240 eV nm)^2/(8*.511E6 eV*1E-2 nm^2)^(2/3) =    24.31011824 eV

#26. 原子尺寸3-D箱中电子的能级。设电子的波长等于箱的尺寸的两倍，因此波长 = 2*V^(1/3) = 2*(1E-29)^(1/3)= 4.30887E-10 m 最小动量 = sqrt(3*(h/(2*V^(1/3))^2) = sqrt(3)*hc/(2*c*V^(1/3)) =sqrt(3)*1240 eV nm/(2*c*(10^-2 nm^2)^(1/3)) = 4984.469966 eV/c b. 基态能量是 E=3*(hc)^2/(8*mc^2*V^(2/3)) = =3*(1240 eV nm)^2/(8*.511E6 eV*1E-2 nm^2)^(2/3) = 24.31011824 eV

#28. Electron with energy 13.6eV in 3-D box.
L=sqrt(3h^2/(8*m*E))=sqrt(3(hc)^2/(8*mc^2*E))
L=sqrt(3*(1240 eV nm)^2/(8*.511E6*13.6 eV)) = 0.288043059 nm

#28 一个能量为13.6 eV的电子在三维盒中。L=sqrt(3h²/(8mE))=sqrt(3(hc)²/(8mc²E)) L=sqrt(3*(1240 eV nm)²/(8*0.511E6*13.6 eV)) = 0.288043059 纳米

#31. Potential well, calculate number of bound states.
For a potential well problem, the energies are just the particle in a box energies minus the depth of the well.
En = - V0 + (nh)^2/(8m(2L)^2)
The coefficient of n^2 is (hc)^2/(8 mc^2 (2L)^2) = (1240 eV nm)^2/(8*.511E6 eV*(.2 nm)^2)
9.403131115 eV
The n=1 state is then -10 + 9.4 eV = -0.6 eV, and the n=2 state is positive,
so there is only one bound state.

#31. 势阱，计算束缚态数目。对于势阱问题，能量就是粒子在盒中的能量减去势阱的深度。En = -V0 + (nh)^2/(8m(2L)^2) 其中n^2的系数是 (hc)^2/(8 mc^2 (2L)^2) = (1240 eV nm)^2/(8*0.511E6 eV*(0.2 nm)^2) = 9.403131115 eV。因此，n=1态为-10 + 9.4 eV = -0.6 eV，而n=2态为正，因此只有1个束缚态。

#33. Estimate ground state energy of electron in nucleus.
Approximating the nucleus as an infinite square well of dimension 5 fermis,
then we can take the wavelength to be on the order of 5 fermis
The momentum is then p=h/wavelength and pc=hc/wavelength
This gives pc= 1240MeV fm/5 fm = 248 MeV
Since this is far above the rest mass energy of 0.511MeV, we see that the energy is approx 248 MeV.

估计电子在核内的基态能量。将核近似为一个边长为5费米的无限深方势阱，那么可以认为波长约为5费米。此时动量为p=h/波长，且pc=hc/波长。这给出pc=1240MeV fm/5 fm=248 MeV。由于这个值远高于静止质量能量0.511MeV，因此能量约等于248 MeV。

尝试开发解决有限深势阱的常规方法

对于无限深势阱，能量由E(n) = (nh)²/(8mL²)给出

n h lm lnm m mev 1 6.63×10⁻³⁴ 2.00×10⁻¹⁰ 0.2 9.11×10⁻³¹ 5.11×10⁵

无限深势阱的能 = 1.50585E-18 焦耳 = 9.403131115 eV

For the finite well, even solution:
For well depth U0 =
1.00E+03 eV 1/beta Effective width Energy for effective width
First approx 9.403131115 0.006209878 0.212419756 8.335711438
8.335711438 0.006206535 0.21241307 8.336236208
8.336236208 0.006206537 0.212413073 8.33623595
8.33623595 0.006206537 0.212413073 8.33623595
8.33623595 0.006206537 0.212413073 8.33623595
8.33623595 0.006206537 0.212413073 8.33623595
8.33623595 0.006206537 0.212413073 8.33623595

对于有限深势阱的偶解：对于势阱深度U0 = 1.00E+03 eV 1/beta 有效宽度能量有效宽度第一近似 9.403131115 0.006209878 0.212419756 8.335711438 8.335711438 0.006206535 0.21241307 8.336236208 8.336236208 0.006206537 0.212413073 8.33623595 8.33623595 0.006206537 0.212413073 8.33623595 8.33623595 0.006206537 0.212413073 8.33623595 8.33623595 0.006206537 0.212413073 8.33623595

Alpha decay of Uranium-238
The alpha from U-238 decay has an energy of 4.27 MeV and a half-life of 4.46 billion years.
The height of the barrier is calculated to be about 35 MeV
The radius of the U-238 nucleus is about 7.4 fermi.
The value of of the potential drops to the alpha energy at distance
r = 7.4 fm *(35/4.27) = 60.6557377 fm
Try to calculate a decay constant based on 5 segments with barrier height equal to the midpoint potential.
Width of segment = 10.65114754
x V(x)-E beta exp(-2*beta*x) net
5.32557377 16.08271687 1.752920345 7.78918E-09
15.97667377 6.80942056 1.140610991 5.29493E-06 4.12431E-14
26.62777377 3.341429468 0.799003391 0.00020139 8.30597E-18
37.27887377 1.526923202 0.540121009 0.003173618 2.636E-20
47.92997377 0.411007099 0.280225166 0.050554351 1.33261E-21

铀-238的α衰变：铀-238衰变产生的α粒子具有4.27 MeV的能量，半衰期为44.6亿年。势垒的高度计算为约35 MeV。铀-238核的半径约为7.4费米。势能值在距离r = 7.4 fm *(35/4.27) = 60.6557377 fm处降至α粒子的能量。尝试基于5个段的宽度，势垒高度等于中间势能的值。段的宽度=10.65114754 x V(x)-E beta exp(-2*beta*x) net 5.32557377 16.08271687 1.752920345 7.78918E-09 15.97667377 6.80942056 1.140610991 5.29493E-06 4.12431E-14 26.62777377 3.341429468 0.799003391 0.00020139 8.30597E-18 37.27887377 1.526923202 0.540121009 0.003173618 2.636E-20 47.92997377 0.411007099 0.280225166 0.050554351 1.33261E-21

With 9.7E20 hits of alpha at radius, this gives a decay constant = 1.72258E-21

And a half-life of 4.02304E+20 sec= 1.27482E+13 years