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

Stefan's constant= sigma= 5.67E-08 watt/m^2K^4
Nuclear distance unit fermi 1.00E-15 m (a femtometer)

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/c² 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 hc hc 1.24E+03 eV nm 速率 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 Stefan 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)

#1. Debroglie wavelengths
a. wavelength of photon with energy 10 eV
wavelength =hc/E = 1.24E-07 m
A typical Rohlf approach would be to use hc = 1240 eV nm to
get wavelength =1240 eV nm/10 eV = 124 nm
b. Electron with kinetic energy 6 MeV
This is clearly relativistic, so use wavelength = h/p = hc/pc
Using the relativistic energy expression E=sqrt((pc)^2+(m0c^2)^2)
you get wavelength = hc/sqrt(E^2 -(m0c^2)^2)
where a KE=6MeV gives E=6.511MeV.
wavelength = 1240/sqrt(6.511E6^2 - .511E6^2)=   0.000191036 nm
c. a neutron with momentum 1 keV/c
pc = 1 keV, so wavelength = h/p=hc/pc=1240eVnm/1000 eV = 1.24 nm
d. a neutrino with energy 1 GeV
Since the mass energy is nearly zero if not identically zero, the neutrino
can be treated the same way as the photon in part a.
wavelength =hc/E =1240 eV nm/1E9 eV =    0.00000124 nm
          = 1.24 fermi

#1. 德布罗意波长 a. 能量为10 eV的光子的波长波长 = hc/E = 1.24×10^-7 m 一种典型的Rohlf方法是使用 hc = 1240 eV nm 来得到波长 = 1240 eV nm / 10 eV = 124 nm b. 动能为6 MeV的电子这显然是相对论性的，所以使用波长 = h/p = hc/pc 使用相对论能量表达式 E = sqrt((pc)^2 + (m0c^2)^2) 你可以得到波长 = hc / sqrt(E^2 - (m0c^2)^2) 其中一个动能为6 MeV 的情况给出 E = 6.511 MeV. 波长 = 1240 / sqrt(6.511E6^2 - 0.511E6^2) = 0.000191036 nm c. 动量为1 keV/c的中子 pc = 1 keV，所以波长 = h/p = hc / pc = 1240 eV nm / 1000 eV = 1.24 nm d. 能量为1 GeV的中微子由于质量能量几乎为零（如果不是完全零），中微子可以像部分a中的光子一样处理。波长 = hc / E = 1240 eV nm / 1E9 eV = 0.00000124 nm = 1.24 fermi

#2. Kinetic energies for particles of given wavelengths
a. Electron with wavelength 0.1 nm
Try to cast in the units eV and nm.
Kinetic energy = p^2/2m for this non-relativistic electron
KE = h^2/(2m*wavelength)^2 = (hc)^2/(2mc^2*wavelength^2)
=(1240eV nm)^2/(2*0.511e6 eV* (0.1nm)^2)= 150.4500978 eV

#2. 波长为0.1 nm的粒子的动能 a. 电子试着将其转换为eV和nm单位。动能 = p²/(2m) 对于这个非相对论电子 KE = h²/(2m*波长)² = (hc)²/(2mc²*波长²) = (1240eV nm)²/(2*0.511e6 eV*(0.1nm)²) = 150.4500978 eV

If the electron were relativistic we could use E=sqrt((pc)^2+(m0c^2)^2) to
get E and then get kinetic energy from E-m0c^2
This wavelength gives pc=hc/wavelength = 12400 eV
Using rest mass energy .511 MeV we get kinetic energy
150.4279564 eV, so to the accuracy of the constants used, they are the same.

如果电子是相对论性的，我们可以用 E=sqrt((pc)^2+(m0c^2)^2) 来计算 E，然后通过 E-m0c^2 得到动能。这个波长给出 pc=hc/波长 = 12400 eV。使用静止质量能量 0.511 MeV，得到动能 150.4279564 eV，因此在所用常数的精度范围内，它们是相同的。

b. photon with wavelength 0.1nm
The photon kinetic energy is given simply by pc=hc/wavelength=1240 eV nm/.1nm
KE= 12.4 keV

b. 波长为0.1纳米的光子，其动能由pc=hc/波长=1240 eV nm/0.1nm KE= 12.4 keV

c. alpha particle with wavelength 1 fm
The mass of the alpha particle =3730 MeV
For a 1 fm wavelength pc=hc/wavelength=1240eV nm/1E-6nm
pc= 1240000000 eV= 1240 MeV
The energy is then = 3930.712404 MeV
The kinetic energy is then = 200.7124036 MeV
Using the non-relativistic expression p^2/2m , even with relativistic mass and
momentum gives = (1240MeV-fm)^2/(2*3730MeV*(1fm)^2) = 206.1126005 MeV

c. 波长为1 fm的α粒子，其质量=3730 MeV。对于波长为1 fm的粒子，pc=hc/波长=1240eV nm/1E-6nm=1240000000 eV=1240 MeV。能量则为=3930.712404 MeV。动能则为=200.7124036 MeV。使用非相对论表达式p²/2m，即使考虑相对论质量与动量，也得到=(1240MeV-fm)²/(2*3730MeV*(1fm)²)=206.1126005 MeV

#5. Electron voltage to be same wavelength as 40 keV x-rays
pc=40keV=hc/wavelength =1.24keV nm/wavelength
wavelength = 0.031 nm
For the electron to have this wavelength, pc=hc/wavelength=1240eV nm/.031 nm
pc = 40000 eV, I.e., pc is the same for the photon and electron
The electron energy is = 512563.1668 eV
So the electron kinetic energy= 1563.166839 eV

#5. 电子电压要与40 keV的X射线具有相同的波长 pc=40keV=hc/波长 =1.24keV nm/波长波长 = 0.031 nm 为了电子具有此波长，pc=hc/波长=1240eV nm/.031 nm pc = 40000 eV，即光子和电子的pc相同。电子能量=512563.1668 eV 所以电子的动能=1563.166839 eV

The non-relativistic KE expression p^2/2m gives 1565.55773
so this is essentially a non-relativistic electron.
Bottom line: 1563 volts of acceleration gives electron 0.1nm wavelength

非相对论动能表达式p²/2m给出1565.55773，因此这基本上是一个非相对论电子。底 line：1563伏特的加速电压使电子具有0.1纳米的波长。

#7. Kinetic energies of electrons, neutrons and photons of wavelength 0.01 nm
Note that the product pc is going to be the same for all three, and that the kinetic energy
can be calculated from pc.
pc= hc/wavelength = 1240 eV nm/0.01 nm = 124000 eV
a. electron
Kinetic energy = E - m0c^2 = sqrt((pc)^2+(m0c^2)^2) -m0c^2
=sqrt(124000^2 + 511000^2) - 511000^2 =   14829.82038 eV

b. neutron
neutron is non-relativistic.
E = p^2/2m = (hc)^2/(2 m0c^2 * wavelength^2) = 1240^2/(2*939.6E6*(.01)^2)
E= 8.182205194 eV

c. photon
For the photon, KE = pc = 124 keV

#8. Debroglie wavelength vs average separation of N2 molecule
a. Taking a temperature of 300K, the average speed of nitrogen molecules
can be obtained from the Maxwell speed distribution
avg speed sqrt(8kT/m*pi)= 476.2883011 m/s

The wavelength can be put in the form wavelength=h/mv=hc/mc^2v
=1240eV nm /(28*931.5E6*476.28)=   9.98185E-11 m

b. The separation between molecules at 300K is about 3.45 nm
so the separation is about   34.56274401 x the wavelength.

#13. Davisson Germer experiment
The Bragg condition for diffraction is n* wavelength = 2d sin theta
The scattering angle is measured to be 50 degrees, so theta = 90-25 = 65 degrees
The electron energy is 54 eV, so we need to calculate wavelength from that.
wavelength = h/p = h/sqrt(2mE) = hc/sqrt(2 mc^2 E)
wavelength = 1240 eV nm/sqrt(2*511000*54) =   0.166916534 nm
Assuming first order, n=1, d = wavelength/(2*sin theta) =    0.092086009 nm

#17. Speed uncertainty of proton located to 1 nm precision.
delta p = hbar/(2 delta x) , delta x = 1 nm
delta v = hbar/(2 m delta x) = c hbar*c/(2 mc^2 delta x)
=3e8 * 197 eV nm/(2* 938E6*1E-9)=   31.50319829 m/s

#20. Electron and proton in atomic-sized volume.
Using the 3D particle in a box expression, the average KE is given by
minimum avg KE = 9 hbar^2/2mL^2
This is for a cube of dimension L, so V=L^3 and L^2=V^(2/3)
For V=10e-30 m^3:    L^2 = 1E-20 m^2
electron avg KE = 5.49269E-18 joules= 34.32929176 eV
proton avg KE= 2.99631E-21 joules= 0.018726937 eV
Note that just the uncertainty principle gives the order of magnitude
of electrons in atoms, whereas it suggests that protons could just wander
in and out of such a volume under the influence of thermal energy.

#7. 波长为0.01 nm的电子、中子和光子的动能注意，pc的乘积对于这三种粒子来说是一样的，而动能可以从pc计算。pc= hc/波长 = 1240 eV nm/0.01 nm = 124000 eV a. 电子动能 = E - m0c^2 = sqrt((pc)^2+(m0c^2)^2) -m0c^2 =sqrt(124000^2 + 511000^2) - 511000^2 = 14829.82038 eV b. 中子中子是非相对论的。E = p^2/2m = (hc)^2

#23. Fractional width of the tau lepton.
From the uncertainty principle in terms of energy,
lambda = hbar/tau, tau given as 3E-13 seconds
lambda = 6.59E-16/3E-13 = 0.002196667 eV
Since the energy is 1.8E9, the fractional width is 2.2E-3/1.8E9 = 1.22 E-12

#23 τ轻子的宽度。根据能量的不确定性原理，λ = ħ/τ，τ给定为3×10⁻¹³秒，λ = 6.59×10⁻¹⁶ / 3×10⁻¹³ = 0.002196667 eV。由于能量是1.8×10⁹ eV，因此宽度的分数是2.2×10⁻³ / 1.8×10⁹ = 1.22×10⁻¹²。

#24. Size of confinement correlated with v/c = 0.5.
If you treat the particle like a standing wave in a one dimensional box
so that the wave has value zero at the walls, this gives wavelength = 2L
Then pc=hc/2L and the particle speed is given by v/c = pc/E=pc/(gamma m0c^2)
For v/c = .5, gamma = 1.15.
The dimension L is than given by L = hc/2pc = hc/2(beta*gamma*m0c^2)
for electron, L=1240MeV fm/(2*.5*1.15*0.511MeV) =    2110.099549 fm
for muon, L=1240MeV fm/(2*.5*1.15*106MeV) =    10.17227235 fm
for charm quark, L=1240MeV fm/(2*.5*1.15*1500MeV) =    0.71884058 fm

粒子受限区域的尺度与 v/c = 0.5 相关。如果将粒子视为一维盒中的驻波，使得波在边界处为零，这给出波长 = 2L。然后 pc = hc/2L，粒子速度由 v/c = pc/E = pc/(γ m0c²) 给出。对于 v/c = 0.5，γ = 1.15。则长度 L 由 L = hc/(2pc) = hc/(2βγm0c²) 给出。对于电子，L = 1240MeV fm/(2×0.5×1.15×0.511MeV) = 2110.099549 fm；对于缪子，L = 1240MeV fm/(2×0.5×1.15×106MeV) = 10.17227235 fm；对于粲夸克，L = 1240MeV fm/(2×0.5×1.15×1500MeV) = 0.71884058 fm。

#25. Kinetic energy for electron confined to nucleus.
If we take the nuclear size to be 2 fermi
delta p c = hbar c/2 delta x,
delta pc = 197 MeV fm/ 4 = 49.25 MeV
for extreme relativistic case like this , pc approx = energy

电子被限制在核内的动能。若将核的尺寸取为2费米，则Δp c = hbar c / 2 Δx，Δp c = 197 MeV fm / 4 = 49.25 MeV（对于这种极端相对论情况），pc近似等于能量。

#28. Width of the K meson.
From the uncertainty principle in energy,
lambda = hbar/tau , tau = 8.9E-11 sec
lambda=6.59E-16 eV s/8.9E-11 = 7.40339E-06 eV

K介子的宽度。根据能量的不确定性原理，lambda = ħ/tau ，tau = 8.9×10⁻¹¹ 秒，lambda=6.59×10⁻¹⁶ eV·s / 8.9×10⁻¹¹ = 7.40339×10⁻⁶ eV