Appendix A Estimates

We came across a number of physical constants that you need to know. Outside of this list, the constants will be given.

1. 1.

   $\mathrm{Giga}={10}^9$, $\mathrm{Mega}={10}^6$, $\mathrm{milli}={10}^{-3}$ so

   $$1\mathrm{meV}={10}^{-3}\mathrm{eV}\mathit{\hspace{1em}}1\mathrm{MeV}={10}^6\mathrm{eV}\mathit{\hspace{1em}}1\mathrm{GeV}={10}^9\mathrm{eV}$$

   (A.1)

2. 2.

   Avogadro’s number $N_A=6\times {10}^{23}$.

3. 3.

   The speed of light $c=3\times {10}^8\mathrm{m}/\mathrm{s}$.

4. 4.

   The speed of sound in air is approximately $c_s\simeq 330\mathrm{m}/\mathrm{s}$.

5. 5.

   A useful unit of volume is liters. One liter is ${(10\mathrm{cm})}^3=1000{\mathrm{cm}}^3$. One mole of an ideal gas at STP has a volume of $22\mathrm{L}$.

6. 6.

   A useful unit of pressure is a bar. One atmosphere is approximately 1 bar. 1 bar is ${10}^5\mathrm{N}/{\mathrm{m}}^2$. A typical scale of energy is $1\mathrm{bar}\times 1\mathrm{L}=100\mathrm{J}$.

7. 7.

   The ideal gas constant is $R=8.32\mathrm{J}/{}^{\circ }\mathrm{K}$:

8. 8.

   The Boltzmann constant $k_B$ you can remember in two ways:

   • •

     The macroscopic way: one Avogadro’s number times $k_B$ is $R$:

     $$N_A{k}_B=R$$

     (A.2)

   • •

     The microscopic way: $kT$ is “one fortieth of an eV at room temperature”, $T=300{}^{\circ }\mathrm{K}$.

     $$k_B=\frac{\frac{1}{40}\mathrm{eV}}{300{}^{\circ }\mathrm{K}}=\frac{0.025\mathrm{eV}}{300{}^{\circ }\mathrm{K}}$$

     (A.3)

9. 9.

   You should remember the proton (and neutron mass) in two ways:

   • •

     The microscopic way: i.e. the rest energy in mega electron volts is

     $$m_p{c}^2\simeq 938\mathrm{MeV}\simeq 1000\mathrm{MeV}\simeq 1\mathrm{GeV}$$

     (A.4)

   • •

     The macroscopic way: an Avogadro’s number of protons weighs a gram. This is the molar mass of the proton:

     $${ℳ}_{\mathrm{ml}}=m_p{N}_A=1\mathrm{g}$$

     (A.5)

   Protons and neutrons and weigh nearly the same thus the mass of one Avogadro’s number of diatomic oxygen weighs $32g$, since there are eight protons and neutrons in one oxygen nucleus, and two such nuclei. The electrons are light (see below) for the mass budget.

   You might want to use either of these methods to evaluate $v_{\mathrm{rms}}$ in atomic hydrogen gas at room temperature

   $$v_{\mathrm{rms}}=\sqrt{\frac{3kT}{m_p}}=c\sqrt{\frac{3kT}{m_p{c}^2}}=(3\times {10}^8\mathrm{m}/\mathrm{s})\times \sqrt{\frac{(1/40)\mathrm{eV}}{938\times {10}^6\mathrm{eV}}}\simeq 2700\mathrm{m}/\mathrm{s}$$

   (A.6)

   Or if you prefer

   $$v_{\mathrm{rms}}=\sqrt{\frac{3k_{B}T}{m}}=\sqrt{\frac{3(N_A{k}_B)T}{N_A{m}_p}}=\sqrt{\frac{3RT}{1g}}=\sqrt{\frac{3\times 8.32\mathrm{J}/{}^{\circ }\mathrm{K}\times 300{}^{\circ }\mathrm{K}}{1g}}\simeq 2700\mathrm{m}/\mathrm{s}$$

   (A.7)

   This is almost a factor of 10 faster than the speed of sound $c_s\simeq 330\mathrm{m}/\mathrm{s}$, because hydrogen is so light.

10. 10.

    You should remember the electron mass in two ways:

    • •

      The microscopic way: the mass is “half an MeV”

      $$m_e{c}^2\simeq 0.511\mathrm{MeV}$$

      (A.8)

    • •

      In comparison to the proton mass:

      $$\frac{m_e}{m_p}\simeq \frac{1}{2000}$$

      (A.9)

11. 11.

    Planck’s constant is needed to convert wavelength to energy

    $$\mathrm{\hslash }c=197\mathrm{eV}\mathrm{nm}$$

    (A.10)

    or using $h=2\pi \mathrm{\hslash }$

    $$hc=1240\mathrm{eV}\mathrm{nm}$$

    (A.11)

    Thus the energy of a photon of yellow light with $\lambda =550\mathrm{nm}$ (emitted by sodium) is

    $$E=\frac{hc}{\lambda }=\frac{1240\mathrm{eV}\mathrm{nm}}{550\mathrm{nm}}\simeq 2.3\mathrm{eV}$$

    (A.12)

    Planck’s constant is also useful for measuring typical de Broglie wavelength at room temperature

    $$\frac{h}{\sqrt{2\pi m_{p}kT}}=\frac{hc}{\sqrt{2\pi (m_p{c}^2)(kT)}}\simeq \frac{1240\mathrm{eVnm}}{\sqrt{2\pi \cdot {10}^9\mathrm{eV}\frac{1}{40}\mathrm{eV}}}\simeq 1.0\mathring{\mathrm{A}}$$

    (A.13)

12. 12.

    A useful unit of distance in atomic physics is angstroms, $1\mathring{\mathrm{A}}=0.1\mathrm{nm}$. The Bohr radius

    $$a_0=0.53\mathring{\mathrm{A}}$$

    (A.14)

    is about half an Angstrom. A typical bond length is normally between 1-5 Bohr Radii. (For $N_2$ the distance between the two nuclei is $1.09\mathring{\mathrm{A}}$)

13. 13.

    An electron volt is a good unit of microscopic energy. Avogadro’s number times $1\mathrm{eV}$ is a good unit of macroscopic chemical energy and is 100 kilo Joules.

    $$N_A\mathrm{eV}\simeq 100\mathrm{kJ}$$

    (A.15)

    This is sometimes called the Faraday constant¹¹ 1 We can quickly find the charge in Coulombs from this relation, $$1eV\simeq 100\mathrm{kJ}/N_A\simeq 1.6\times {10}^{-19}\mathrm{J}\mathit{\hspace{1em}\hspace{1em}}\text{and so}\mathit{\hspace{1em}\hspace{1em}}1e=1.6\times {10}^{-19}\mathrm{C}.$$ (A.16) . An explosion involves roughly an Avogadro’s number of atomic transitions, with each atomic transition releasing about an electron volt of energy, for approximately $100\mathrm{kJ}$ of energy per mole. Burning a mole of gasoline gives roughly this amount of energy.

14. 14.

    The Bohr model provides a lot of estimates of the microscopic world.

    In the Bohr model the radius of the lowest electron orbit is $a_0=0.53\mathring{\mathrm{A}}$, and the angular momentum of the lowest orbit is discrete, $L=p{a}_0=\mathrm{\hslash }$. So the momentum is

    $$p=\frac{\mathrm{\hslash }}{a_0}$$

    (A.17)

    reflecting the uncertainty principle. In the lowest orbit Bohr orbit the kinetic energy is half of the potential energy in magnitude²² 2 This follows from Newton’s Law for a circular orbit for an electron attracted to the proton via the Coulomb force: $$\frac{m{v}^2}{r}=\frac{e^2}{4\pi {ϵ}_0{r}^2}.$$ Multiplying this equation by $r/2$ gives $$\frac{1}{2}m{v}^2=\frac{1}{2}\frac{e^2}{4\pi {ϵ}_{0}r}.$$ . The potential energy is negative reflecting the positive charge $+e$ of the proton and negative charge $-e$ of the electron. The potential energy of the orbiting electron is

    $$\mathrm{PE}=-\frac{e^2}{4\pi {ϵ}_0{a}_0}$$

    (A.18)

    The binding energy of the electron to the proton

    	$E=$	$\mathrm{KE}+\mathrm{PE}$		(A.19)
    	$=$	${\displaystyle \frac{p^2}{2m}}+\left(-{\displaystyle \frac{e^2}{4\pi {ϵ}_0{a}_0}}\right)$		(A.20)
    	$=$	$-13.6\mathrm{eV}$		(A.21)

    Thus we have for the Hydrogen atom where $p=\mathrm{\hslash }/a_0$ and $\mathrm{KE}=p^2/2m$

    $$\mathrm{KE}=\frac{1}{2}|\mathrm{PE}|=|E|$$

    (A.22)

    $$\boxed{\frac{{\mathrm{\hslash }}^2}{2m_e{a}_0^2}=\frac{1}{2}\left(\frac{e^2}{4\pi {ϵ}_0{a}_0}\right)\simeq 13.6\mathrm{eV}}$$

    (A.23)

    $13.6\mathrm{eV}$ is the Rydberg constant.

    For the ground states of a quantum mechanical system it is generically the case that

    $$\mathrm{KE}\sim |\mathrm{PE}|\sim |E|$$

    (A.24)

    If the system size is of order $L$ then $p\sim \mathrm{\hslash }/L$ and the kinetic energy is ${\mathrm{\hslash }}^2/2m{L}^2$, and often this a good estimates for $|PE|$ and $|E|$. So the intuition from the Bohr model is often a good guess for more general quantum mechanical systems.

    These facts give another way to estimate the thermal de Broglie wavelength of a proton at room temperature, by inserting $a_0$ and the electron mass, and noting that $h=2\pi \mathrm{\hslash }$:

    $${\lambda }_{\mathrm{th}}\equiv \frac{(2\pi \mathrm{\hslash })}{\sqrt{2\pi m_{p}kT}}=2\pi a_0\sqrt{\frac{{\mathrm{\hslash }}^2}{2m_e{a}_0^2}\left(\frac{1}{k_{B}T}\right)}\sqrt{\frac{m_e}{\pi m_p}}=2\pi (0.5\mathring{\mathrm{A}})\sqrt{\frac{13.6\mathrm{eV}}{0.025\mathrm{eV}}}\sqrt{\frac{1}{\pi 2000}}\simeq 1\mathring{\mathrm{A}}$$

    (A.25)