Problem Set 1

2.1

(a)

How many atoms are there in a yoctomole?

Yocto is the SI prefix for $10^{-24}$ . So there are $\frac{N_a}{\si{\mol}} \cdot 10^{-24} = \frac{6.022 \cdot 10^{23}}{\si{\mol}} \cdot 10^{-24} = 0.6022$ atoms in a yoctomole.

(b)

How many seconds are there in a nanocentury? Is the value near that of any important constants?

Nano is the SI prefix for $10^{-9}$ .

$\num{1e-9} \si{century} \cdot \frac{100 \si{years}}{1 \si{century}} \cdot \frac{365 \si{days}}{1 \si{year}} \cdot \frac{24 \si{hours}}{1 \si{day}} \cdot \frac{60 \si{minutes}}{1 \si{hour}} \cdot \frac{60 \si{\s}}{1 \si{minute}} = 3.15 \si{\s}$

This is relatively close to $\pi$ .

2.2

A large data storage system holds on the order of a petabyte. How tall would a 1 petabyte stack of CDs be? How does that compare to the height of a tall building?

CDs are nominally $1.2 \si{\mm}$ thick, and store 650 megabytes. (Some actually store more because they write data in regions the Orange Book standard reserves for manufacturing slop.)

$10^{15} \si{bytes} \cdot \frac{\num{1.2e-3} \si{\m}}{\num{6.5e8} \si{bytes}} \cdot \frac{\si{km}}{10^3 \si{m}} = 1.8 \si{\km}$

Currently the tallest building in the world is $0.8 \si{km}$ (the Burj Khalifa). The Jeddah Tower, currently under construction, is intended to be $1 \si{km}$ . So for the foreseeable future the stack of CDs would be roughly twice as tall as any other human-built structure.

2.3

If all the atoms in our universe were used to write an enormous binary number, using one atom per bit, what would that number be (converted to base 10)?

There are about $10^{80}$ atoms in the observable universe. So if each of these was used to encode one bit, we’d have $10^{80}$ bits of information. If all of this was used to store one big unsigned integer, the largest representable value would be $2^{10^{80}} - 1$ (not that the 1 makes any noticeable difference). Every decimal digit encodes $\log_2(10) \approx 3.32$ bits of information, so $10^{80}$ bits is equivalent to $10^{80} / \log_2(10) \approx \num{3e79}$ decimal digits of information. Thus the largest representable unsigned integer can also be written as $10^\num{3e79}$ .

2.4

Compare the gravitational acceleration due to the mass of the Earth at its surface to that produced by a 1 kg mass at a distance of 1 m. Express their ratio in decibels.

The acceleration due to gravity at the surface of the Earth is $9.8\si{m.s^{-2}}$ , while that of a $1\si{kg}$ mass at $1\si{m}$ is $G \cdot 1\si{kg} / 1\si{m^2} = \num{6.67e-11} \si{m.s^{-2}}$ . In decibels this is

$20 \cdot \log_{10} \frac{9.8 \si{m.s^{-2}}}{\num{6.67e-11} \si{m.s^{-2}}} = 223 \si{db}$

2.5

(a)

Approximately estimate the chemical energy in a ton of TNT. You can assume that nitrogen is the primary component; think about what kind of energy is released in a chemical reaction, where it is stored, and how much there is.

TNT is a chemical explosive, meaning its potential energy is stored in chemical bonds. Chemical bonds are held together by the electromagnetic force. Covalent bonds in particular result from overlapping orbitals that enable multiple atoms to share electrons. I don’t know what compounds result from the combustion of TNT, but clearly the atoms must end up in lower energy states on average. So for a rough estimate, I’ll compute the number of nitrogen atoms per ton of TNT, and assume we free around 1 eV per atom during combustion, since this is a typical amount of energy for an atomic excitation.

$1 \si{ton} \cdot \frac{\num{907.18e3} \si{g}}{\si{ton}} \cdot \frac{\si{mol}}{14 \si{g}} \cdot \frac{\num{6.022e23} \si{atoms}}{\si{mol}} \cdot \frac{1 \si{eV}}{\si{atom}} \cdot \frac{\num{1.6e-19} \si{J}}{\si{eV}} = \num{6e9} \si{J}$

This is off by a factor of about 1.5, which isn’t so bad for not knowing what the reaction is.

(b)

Estimate how much uranium would be needed to make a nuclear explosion equal to the energy in a chemical explosion in 10,000 tons of TNT (once again, think about where the energy is stored).

Nuclear explosions utilize nuclear energy instead of atomic (electromagnetic) energy. So we’re dealing with a typical energy of $10^6 \si{eV}$ rather than $1 \si{eV}$ . On top of that, we’re getting about that much energy per proton or neutron, rather than per atom. Most uranium has 238 nucleons, but fissionable uranium has only 235. However we’re penalized for atomic mass: uranium is much heavier than nitrogen so there are less atoms of it per ton. So overall uranium should be about $10^6 \cdot 235 \cdot 14/235 = \num{1.4e7}$ times as potent as TNT per unit mass. Thus $10^4$ tons of TNT is roughly equivalent to $10^4 / (\num{1.4e7}) = \num{7e-4}$ tons of fissionable uranium, which is about 650 grams.

(c)

Compare this to the rest mass energy $E = mc^2$ of that amount of material (Chapter 15), which gives the maximum amount of energy that could be liberated from it.

$E = mc^2 = 0.65 \si{kg} \cdot (\num{3e8} \si{m/s})^2 = \num{5.8e16} \si{J}$

This is roughly one quarter as powerful as Tsar Bomba, the largest single bomb ever built, which had a 50 megaton = $\num{2e17} \si{J}$ yield. It weighted 27 megagrams, though not all of that was fissile material. Still, if we had that much antimatter, we could get

$E = mc^2 = 2 \cdot \num{27e6} \si{kg} \cdot (\num{3e8} \si{m/s})^2 = \num{5e24} \si{J}$

I incude the factor of 2 since the antimatter would convert an equivalent amount of matter to energy on contact.

Let’s compare this to the gravitational binding energy of the planet, since this is the final frontier of any serious arms race within a uniplanetary species. We can calculate this by finding the energy required to remove a thin shell of matter from the surface of a sphere (moving it infinitely far away), and integrating it from the center of the Earth to the surface. For each shell, the spherical ball of matter inside behaves like a point mass, so the binding energy is just $G M m / r$ . Assume a constant density $\rho$ , so that the mass of a sphere of radius $r$ is $(4/3) \pi r^3 \rho$ , and the mass of the shell on top of it is $4 \pi r^2 \rho \mathrm{d}r$ . Thus the binding energy of the shell is $G ((4/3) \pi r^3 \rho) (4 \pi r^2 \rho \mathrm{d}r) / r$ . The total energy needed to unbind the Earth is

$\begin{align*} E &= \int_0^R \frac{16}{3} G \pi^2 \rho^2 r^4 \mathrm{d}r \\ &= \frac{16}{3} G \pi^2 \rho^2 \left[ \frac{1}{5} r^5 \right]_0^R \\ &= \frac{16}{15} G \pi^2 \rho^2 R^5 \\ \end{align*}$

Since $\rho$ must be $3 M/(4 \pi R^3)$ , we have

$\begin{align*} E &= \frac{3 G M^2}{5 R} \\ &= \frac{3 \cdot (\num{6.67e-11} \si{m^3.kg^{-1}.s^{-2}}) \cdot (\num{6e24} \si{kg})^2}{5 \cdot (\num{6.4e6} \si{m})} \\ &= \num{2e32} \si{J} \end{align*}$

So you’d need a lot more than an antimatter Tsar Bomba to turn the Earth to dust. In fact, you’d need about 3 exagrams of the stuff. Wikipedia tells me this is about the mass of all coal deposits on Earth. Luckily, we can’t currently produce (and catch) more than billions of antiprotons, which puts us about 14 orders of magnitude away from creating a single gram. (Coincidentally this also makes it the world’s most valuable substance by mass.)

2.6

(a)

What is the approximate de Broglie wavelength of a thrown baseball?

A baseball weighs $1.45^{-1} \si{kg}$ . A major league fastball is around 100 miles per hour, so 160 km per hour, but I’ll just go with 100 km per hour since we’re not all professional pitchers.

$\begin{align*} \lambda &= \frac{h}{p} \\ &= \frac{\num{6.626e-34} \si{J.s}}{\num{1.45e-1} \si{kg} \cdot 10^2 \si{km/hour}} \cdot \frac{\si{hour}}{60 \si{minutes}} \cdot \frac{\si{minute}}{60 \si{s}} \\ &= \num{1.3e-35} \si{m} \end{align*}$

(b)

Of a molecule of nitrogen gas at room temperature and pressure? (This requires either the result of Section 3.4.2, or dimensional analysis.)

By the equipartition theorem, there is $\frac{1}{2} k T$ energy in the x, y, and z components of the molecule’s velocity. We know $E = \frac{1}{2} m v^2$ , and $p = m v$ , so $p = \sqrt{m k T}$ . Since the x, y, and z components add in quadrature, a typical total momentum will be $\sqrt{3 m k T}$ . The mass of a molecule of N2 is

$2 \cdot \frac{14 \si{g}}{\si{mol}} \cdot \frac{\si{mol}}{\num{6.022e23}} = \num{4.6e-23} \si{g} = \num{4.6e-26} \si{kg}$

Thus

$\begin{align*} \lambda &= \frac{h}{p} \\ &= \frac{h}{\sqrt{3 m k T}} \\ &= \frac{\num{6.626e-34} \si{J.s}}{\sqrt{3 \cdot \num{4.6e-26} \si{kg} \cdot \num{1.38e-23} \si{J/K} \cdot 300 \si{K}}} \\ &= \num{2.7e-11} \si{m} \end{align*}$

(c)

What is the typical distance between the molecules in this gas?

The typical distance can be estimated based on the number of particles per volume. If the volume is $V$ , then we can arrange $n$ particles on a cubic lattice with a spacing of $(V / n)^{1/3}$ . By the ideal gas law, $V/n = RT/P$ , so the typical distance is

$\begin{align*} \left( \frac{V}{n} \right) ^ \frac{1}{3} &= \left( \frac{R T}{P} \right) ^ \frac{1}{3} \\ &= \left( \frac{8.3 \si{J/mol/K \cdot 300 K}}{10^5 \si{Pa}} \cdot \frac{\si{mol}}{\num{6.022e23}} \right) ^ \frac{1}{3} \\ &= \num{3.5e-9} \si{m} \end{align*}$

(d)

If the volume of the gas is kept constant as it is cooled, at what temperature does the wavelength become comparable to the distance between the molecules?

The typical spacing depends on the volume of the gas and number of molecules, both of which are fixed in this scenario. So we just need to solve $\num{3.5e-9} \si{m} = \lambda = h / \sqrt{3 m k T}$ for $T$ .

$\begin{align*} T &= \frac{h^2}{3 m k \lambda^2} \\ &= \frac{(\num{6.626e-34} \si{J.s})^2}{3 \cdot \num{4.6e-26} \si{kg} \cdot \num{1.38e-23} \si{J/K} \cdot (\num{3.5e-9} \si{m})^2} \\ &= 0.02 K \end{align*}$

(2.7)

(a)

The potential energy of a mass m a distance r from a mass $M$ is $−GMm/r$ . What is the escape velocity required to climb out of that potential?

Assuming small and slow enough masses that relativistic corrections are negligible, we just need to solve

$E = \frac{1}{2} m v^2 = \frac{G M m}{r} \implies v = \sqrt{\frac{2 G M}{r}}$

(b)

Since nothing can travel faster than the speed of light (Chapter 15), what is the radius within which nothing can escape from the mass?

$c = \sqrt{\frac{2 G M}{r}} \implies r = \frac{2 G M}{c^2}$

This derivation is very much not rigorous, since neither the gravitational potential nor kinetic energy is expressed relativistically (in fact there’s no such thing as a gravitational potential in general relativity, since gravity isn’t a force). But remarkably we do end up with the real Schwarzschild radius.

(c)

If the rest energy of a mass $M$ is converted into a photon, what is its wavelength?

$E = M c^2 = \frac{h c}{\lambda} \implies \lambda = \frac{h}{M c}$

(d)

For what mass does its equivalent wavelength equal the size within which light cannot escape?

$\frac{h}{M c} = \frac{2 G M}{c^2} \implies M = \sqrt{\frac{h c}{2 G}} \approx \num{3.9e-8} \si{kg}$

(e)

What is the corresponding size?

Plugging the above formula for $M$ into $\lambda = h/(M c)$ , we find that

$\lambda = \sqrt{\frac{2 h G}{c^3}} \approx \num{5.7e-35} \si{m}$

(f)

What is the energy?

$E = \frac{hc}{\lambda} = \sqrt{\frac{h c^5}{2 G}} \approx \num{3.5e9} \si{J}$

(g)

What is the period?

$\tau = 1/f$ and $\lambda = c / f = c \tau$ , so

$\tau = \lambda / c = \sqrt{\frac{2 h G}{c ^ 5}} \approx \num{1.9e-43} \si{s}$

2.8

Consider a pyramid of height H and a square base of side length L. A sphere is placed so that its center is at the center of the square at the base of the pyramid, and so that it is tangent to all of the edges of the pyramid (intersecting each edge at just one point).

(a)

How high is the pyramid in terms of L?

To fit within the square base, the radius of the sphere must be $L / 2$ . Now imagine rotating the square base 90 degrees about one of its diagonals. Since the axis of rotation goes through the center of the sphere, the edges of the base remain tangent to the sphere. So the height of the pyramid is the same as the distance from the center of the sphere to any corner of the square, namely $L / \sqrt{2}$ . (This could also be calculated using triangle similarities, but that’s a lot more tedious.)

(b)

What is the volume of the space common to the sphere and the pyramid?

The sphere is tangent to three points on each wall of the pyramid (the midpoints of each edge). The intersection of each of these planes with the sphere is a circle, and the spherical caps on top of these circles are the portions outside the pyramid. So we can compute the volume of the intersection of the sphere and the pyramid by subtracting the volume of these caps from the volume of the (half) sphere.

The volume of the caps can be computed via an integral over perpendicular distance from the planes, but to do so we need to know how far these planes are from the center of the sphere. If we align two of the base square’s corners with the X and Y axes (so the base square is the set of points with L1 norm of $L/\sqrt{2}$ ), and the tip of the pyramid along Z, then the normal vector of the planar wall in the first octant is $(1/\sqrt{3}, 1/\sqrt{3}, 1/\sqrt{3})$ . The vector from the center of the sphere to the X base corner is $(L/\sqrt{2}, 0, 0)$ , so the distance from the center of the sphere to the plane is $(L/\sqrt{2}, 0, 0) \cdot (1/\sqrt{3}, 1/\sqrt{3}, 1/\sqrt{3}) = L/\sqrt{6}$ . (Again this could also be derived from triangle similarities.)

Thus the volume of a single cap is

$\begin{align*} V_\text{cap} &= \int_\frac{L}{\sqrt{6}}^\frac{L}{2} \pi r^2 \mathrm{d}x \\ &= \int_\frac{L}{\sqrt{6}}^\frac{L}{2} \pi \left( \frac{L^2}{4} - x^2 \right) \mathrm{d}x \\ &= \left[ \frac{\pi L^2}{4} x \right]_\frac{L}{\sqrt{6}}^\frac{L}{2} - \left[ \frac{\pi x^3}{3} \right]_\frac{L}{\sqrt{6}}^\frac{L}{2} \\ &= \pi L^3 \left( \frac{1}{12} - \frac{7}{36 \sqrt{6}}\right) \end{align*}$

And the volume shared by the sphere and pyramid is

$\frac{1}{2} \cdot V_\text{sphere} - 4 \cdot V_\text{cap} = \pi L^3 \left( \frac{7}{9 \sqrt{6}} - \frac{1}{4} \right)$