Bilingual edition: English is preserved and Chinese follows each unit. Terminology uses the confirmed v20260916 glossary; automated semantic review remains traceable.
中英双语版：英文原文完整保留，中文紧随对应单元；术语采用 v20260916 确认表，自动语义审校结果可追溯。
Homework 1, Physics 3401, Chapter 1, #14,18,22,23,31,32,34,36  Rohlf
作业1，物理3401，第一章，第14、18、22、23、31、32、34、36题，Rohlf

Physical constants

物理常数

Avogadro's number  A 6.02E+23
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

阿伏伽德罗常数 A 6.02E+23 比尔兹曼常数 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/c^2

Problem 1: Calculate atom size for gold
 density of gold  19300 kg/m^3
 atomic mass of gold  197

计算金的原子尺寸，已知金的密度为19300 kg/m³，金的原子质量为197。

 atomic volume = molar mass/(density*Avogadro's number)
 atomic volume = (.197 kg)/(19300 kg/m^3 * 6.02x10^23 atoms/mole)
 atomic volume = 1.69556E-29 m^3/atom
 Taking the cube root of this volume gives d =   2.57E-10 m= 0.257 nm
 Treating volume as a sphere gives d=   3.19E-10 m= 0.319 nm

原子体积 = 摩尔质量/(密度×阿伏伽德罗常数) 原子体积 = 0.197 kg/(19300 kg/m³ × 6.02×10²³ atoms/mole) 原子体积 = 1.69556E-29 m³/atom 对这个体积开立方根得到 d = 2.57E-10 m= 0.257 nm 将体积视为球体得到 d= 3.19E-10 m= 0.319 nm

 Twice covalent radius from periodic table:   0.268 nm

元素周期表中的两倍共价半径：0.268 纳米

Problem 2:Calculate density of aluminum if its diameter is the nominal 0.3nm .
 atomic mass of aluminum = 27
 density = molar mass/(atomic volume*Avogadro's number)
 density =   (27/((0.3e-9m)^3*6.02e23)
 density =   1666.67 kg/m^3
 Actual density = 2700 kg/m^3   density =   3423.548659 kg/m^3
    using covalent diameter 0.236nm from periodic table.
Problems from Rohlf:
14. A. Kinetic energy of hydrogen orbit, assuming circular.
Potential energy -ke^2/r
Force relationship, centripetal acceleration:  mv^2/r = ke^2/r^2
Kinetic energy mv^2/2 =(r/2)*(mv^2/r)= ke^2/(2r) = -PE/2

问题2：如果铝的直径为名义值0.3nm，计算其密度。铝的原子质量为27。密度=摩尔质量/(原子体积*阿伏伽德罗常数) 密度=27/((0.3e-9m)^3*6.02e23) 密度=1666.67 kg/m^3 实际密度=2700 kg/m^3 密度=3423.548659 kg/m^3 使用周期表中给出的共价直径0.236nm。Rohlf的问题：14. A. 氢轨道的动能，假设为圆形。势能-ke^2/r 力关系，向心加速度：mv^2/r = ke^2/r^2 动能mv^2/2 = (r/2)*(mv^2/r) = ke^2/(2r) = -PE/2

b. Size of orbit
  -13.6eV = -1.44 eV nm/(2*r)
r=  0.0529 nm

轨道大小 -13.6 eV = -1.44 eV nm/(2*r) r= 0.0529 nm

c. Speed of electron in orbit
KE=mv^2/2=13.6 eV
v=sqrt(2*13.6eV*1.6e-19J/eV/(9.11e-31kg))=   2.19E+06 m/s= 0.00729 c

c. 原子轨道中电子的速率 KE=mv²/2=13.6 eV v=√(2*13.6eV*1.6e-19J/eV/(9.11e-31kg))= 2.19E+06 m/s= 0.00729 c

18. A. Gravity negligible because it is down to about 10^-39 x electric force.
b. The electron velocity in Ex 1-4 is given to be 2.9e7 m/s, and the distance specified
is L=0.050 meters.
The time of transit is then L/v = 0.05/2.9e7=   1.72414E-09 sec
The gravitational deflection is then 0.5gt^2 =0.5*9.8*(1.72e-9)^2=    1.44962E-17 m
So the gravitational deflection is quite negligible, being on the order of 1/100 the nuclear size.

18. A. 由于重力大约是10^-39倍的电力，因此可以忽略不计。b. 在第1-4题中，电子的速度已给定为2.9e7 m/s，指定的距离是L=0.050米。则通过时间是L/v = 0.05/2.9e7= 1.72414E-09秒。重力偏移量则为0.5gt^2 =0.5*9.8*(1.72e-9)^2= 1.44962E-17 m。因此，重力偏移量非常微小，大约是核尺寸的1/100。

22. Radius of curvature of 10keV electrons in 1 milliTesla field.
r= mv/qB for charge in magnetic field.
v=sqrt(2E/m)=sqrt(2*1e4 eV*1.6e-19J/eV/9.11e-31kg)=    5.9267E+07 m/s
r= (9.11e-31 kg*5.927e7m/s/(1.6e-19C*1e-3T)=    0.337 m
 
 
 

22. 10keV电子在1毫特斯拉磁场中的曲率半径。r= mv/qB 对于磁场中的电荷。v=sqrt(2E/m)=sqrt(2*1e4 eV*1.6e-19J/eV/9.11e-31kg)= 5.9267E+07 m/s r= (9.11e-31 kg*5.927e7m/s/(1.6e-19C*1e-3T)= 0.337 m

23.  Electron at 10^6 m/s has same radius as proton; find speed of proton.

速率为10^6 m/s的电子具有与质子相同的轨道半径；求质子的速率。

  m(p)v = m(e)*10^6
  10^6/ 1836 = 544.66 m/s
 

31. Compare mass energy of mosquito to kinetic energy of 747 jumbo jet.

将蚊子的静质量能量与747大型喷气式飞机的动能进行比较。

 Approximate mosquito as cylinder of water(or blood) about 0.5mm by 3mm.
 Density of water 1 gm/cm^3
 This gives mass pi*r^2*L =   5.89049E-07 kg for mosquito
 The mass energy is obtained from mc^2 =   53014376029 joules for mosquito

将蚊子近似为一个约0.5毫米×3毫米的水（或血液）柱。水的密度为1克/立方厘米，这给出蚊子的质量为πr²L = 5.89049E-07千克。蚊子的质能由mc² = 53014376029焦耳获得。

 Guessing that a 747 is 50 meters long with a diameter of 10 meters and a density of 0.2  gm/cm^3
 The mass projects to  785398.1634 kg

假设一架747飞机长50米，直径为10米，密度为0.2克/立方厘米，其质量约为785398.1634千克。

 For a cruising speed of 600 mi/hr =   268.224 m/s

以600英里/小时的巡航速度 = 268.224米/秒

 The kinetic energy mv^2/2 =  28252387571
 So I come out with roughly comparable energies for the mosquito and jumbo jet, from this point of view.
 

动能 mv²/2 = 28252387571 所以从这个角度看，蚊子和大型喷气式飞机的能大致相当。

32. One ton of TNT =   4.00E+09  joules according to Rohlf, from the chemical
energy release of the explosion.

32. 根据Rohlf的数据，1吨TNT相当于4.00×10⁹焦耳的化学能释放。

U235 releases 200 MeV of energy from one nucleus.     this is = 3.2E-11 joules
U235 has an atomic mass of 235.04 * 1.66x10^-27 =    3.90166E-25 kg

U235释放出每个原子核的200 MeV能量。这等于3.2×10^-11焦耳。U235的原子质量为235.04 × 1.66×10^-27 = 3.90166×10^-25 kg

The required mass of uranium projected from these numbers is     4.87708E-05 kg per ton of TNT

根据这些数值计算出的铀的质量为每吨TNT 4.87708E-05 kg

For a megaton, the mass of U235 required would project to    48.7708 kilograms
 

对于1兆吨而言，所需铀-235的质量约为48.7708千克

34. Find binding energy of deuteron and U238.
 proton mass 0.9383 GeV
 neutron mass 0.93957 GeV binding energy of deuterium=  0.00227 GeV
 deuteron mass 1.8756 GeV   2.27 MeV
 U238 mass 221.697 GeV
 mass unit= 1.66E-27 kg
 92 protons 86.3236 GeV
 146 neutrons 137.17722 GeV
 sum of nucleons 223.50082 GeV binding energy of U238 =  1.80382 GeV
     which = 7.57907563 MeV/nucleon
 

34. 计算氘核和铀-238的结合能。质子质量0.9383 GeV，中子质量0.93957 GeV，氘核的结合能为0.00227 GeV，氘核质量1.8756 GeV，铀-238质量221.697 GeV，质量单位=1.66×10<sup>-27</sup> kg，92个质子总质量86.3236 GeV，146个中子总质量137.17722 GeV，核子总质量223.50082 GeV，铀-238的结合能为1.80382 GeV，即每核子7.57907563 MeV。

36. Fusion energy release = 25 MeV/fusion to yield 4 x 10^26 watts. What mass of hydrogen?

36. 核融合释放的能源为每次融合释放25 MeV，相当于4×10^26瓦特。需要多少质量的氢？

 25 MeV= 4E-12 joules
So for 4.00E+26 watts you need 1E+38 fusions/s or 4E38 hydrogens

25 MeV等于4×10⁻¹²焦耳。因此，对于4.00×10²⁶瓦的功率，需要1×10³⁸次融合/秒或4×10³⁸个氢原子。

For proton mass   1.67E-27 kg, this translates to a mass   6.688E+11 kg/s

对于质子质量1.67×10^-27 kg，这相当于质量6.688×10^11 kg/s。