The cross section for $0.01-\mathrm{MeV}$ neutrons to be captured by uranium 238 is $0.7$ barn. Two grams of ${ }^{238} \mathrm{U}$ is exposed to a flux of $3 \times 10^9$ neutrons per square meter (with energy $0.01 \mathrm{MeV}$ ). How many atoms of ${ }^{239} \mathrm{U}$ will be produced?

To find the mass of the neutron, Chadwick used the reaction

$$\alpha+{ }^{11} \mathrm{~B} \rightarrow \mathrm{n}+{ }^{14} \mathrm{~N}$$

The best available values for the masses of the three nuclei were $m_\alpha=4.00106 \mathrm{u}, m_{\mathrm{B}}=11.00825 \mathrm{u}$, and $m_{\mathrm{N}}=14.0042 \mathrm{u}$. The kinetic energies Chadwick measured as $K_\alpha=5.26 \mathrm{MeV}, K_{\mathrm{B}}=0, K_{\mathrm{n}}=3.26 \mathrm{MeV}$, and $K_{\mathrm{N}}=0.57 \mathrm{MeV}$. Using these data, find Chadwick's value for the mass of the neutron, and compare with the current value.

Use the data of Appendix $D$ to find the binding energies of ${ }^4 \mathrm{He},{ }^{40} \mathrm{Ca}$, and ${ }^{120} \mathrm{Sn}$.

Before two nuclei can react, they must come within $1 \mathrm{fm}$ or so of one another, and because of their Coulomb repulsion, this requires a lot of energy. (a) Calculate the energy needed to bring a proton from infinity to a separation of $1 \mathrm{fm}$ from an alpha particle. $(1 \mathrm{fm}=$ distance from surface to surface. $)$ (b) Do the same for an alpha particle and a uranium nucleus. [Use the above eqution to estimate the radii involved.]

Apply Appendix D to find the chemical symbols and the natural percent abundances of the isotopes with (a) $Z=2, \quad A=3 ; \quad$ (b) $Z=6, \quad A=13$; (c) $Z=26, A=56$; (d) $Z=82, A=206$.Appendix D to find the chemical symbols and the half-lives of the four radioactive nuclei with (a) $Z=1, \quad A=3 ;$ (b) $Z=6, \quad A=14$; (c) $Z=19, A=40$; (d) $Z=94, A=244$.

When Rutherford first observed the reaction , he assumed that the alpha particle simply knocked a proton out of the ${ }^{14} \mathrm{~N}$ in the process $\alpha+{ }^{14} \mathrm{~N} \rightarrow \alpha+\mathrm{p}+{ }^{13} \mathrm{C}$. Given that the incident alpha particles had about $5.6 \mathrm{MeV}$, was this reaction possible?

Suppose two energy levels of the helium atom, in both of which the two electrons' spins are antiparallel (so that the total spin is zero, and the spins can be ignored) and one of the electrons is in the lowest ( $1 s$ ) orbital. In the upper level the second electron has $I=2$; in the lower level the second electron has $l=1$. The atom is placed in a magnetic field, and (as described in Section 9.4) the upper level splits into five equally spaced sublevels and the lower into three sublevels (with the same spacing). (a) Sketch the resulting levels. (b) There are, in principle, 15 different possible transitions from the upper $(I=2)$ level to the lower $(l=1)$ level. Show that because the sublevels all have the same spacing, there are actually only seven distinct energy differences. (c) The selection rules for these transitions are $\Delta l=\pm 1$ and $\Delta m=0$ or $\pm 1$; that is, only transitions that satisfy these rules are allowed. Indicate all of the allowed transitions on your energy level diagram. (d) How many distinct photon frequencies will result from allowed transitions between the two levels? This is the normal Zeeman effect described in the above section.

Use the above equation and the data of Appendix D to predict $\Delta K$, the kinetic energy released in the reaction $\mathrm{p}+{ }^7 \mathrm{Li} \rightarrow \alpha+\alpha$.

In most nuclear reactions the total nucleon number is conserved; that is, $\sum A_{\text {initial }}=\sum A_{\text {final }}$, where $A=Z+N$ for each nucleus, as usual. Show that, except in reactions involving $\beta$ decay, both $\sum Z$ and $\sum N$ are separately conserved. [Hint: Total charge is always conserved, and except in $\beta$ decay, all the charge is carried by protons.]

Consider the $\alpha$ decay of a nucleus $\mathrm{X}$ at rest, $\mathrm{X} \rightarrow \mathrm{Y}+\alpha$, where $\mathrm{Y}$ denotes the offspring nucleus. (a) Prove that because the $\alpha$ is much lighter than Y, almost all the available kinetic energy goes to the $\alpha$; specifically, prove that

$$K_\alpha=\frac{m_{\mathrm{Y}}}{m_\alpha} K_{\mathrm{Y}}$$

[Hint: Use conservation of momentum and treat all particles nonrelativistically.] (b) In the $\alpha$ decay of a typical naturally radioactive nucleus, what percentage of the available kinetic energy does the alpha particle take? (c) Assuming that $\beta$ decay has the form $\mathrm{X} \rightarrow \mathrm{Y}+\mathrm{e}$, use the relation corresponding to (17.86) to find what percentage of the available kinetic energy goes to the electron. (Your answer here is really an upper bound, since some energy can go to the neutrino. It is also only approximate since the electron can be fairly relativistic.)

Use the data in Appendix D to write down the whole of the radioactive series that begins with ${ }^{235} \mathrm{U}$. Draw the series on a plot of $Z$ against $N$ as in the above figure.

In the forbidden region the wave-function amplitude decays as $\psi(x)=\psi_0 e^{-x / \lambda}$. Define $P$ as the ratio of the probabilities of finding the electron at $x$ and at $0: P=\left|\psi(x) / \psi_0\right|^2$. (a) Derive a formula for $x$ in terms of $P$. (b) For a tunneling depth of $\lambda=0.1 \mathrm{~nm}$ (typical of wave-function decay outside a metal), what is the distance $x$ at which $P=10^{-10} ? 10^{-20}$ ?

What is the longest wavelength radiation that can produce a current in a photovoltaic cell that consists of a silicon $p n$ diode with band gap of $1.1 \mathrm{eV}$ ?

Use the data in Appendix $\mathrm{D}$ to write down the whole of the radioactive series that begins with ${ }^{238} \mathrm{U}$. Draw the series on a plot of $Z$ against $N$ as in the above figure.