## Problems & Exercises

(a) $\text{charge:}\left(+1\right)+\left(-1\right)=0;\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}\text{electron family number:}\phantom{\rule{0.25em}{0ex}}\left(+1\right)+\left(-1\right)=0;\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}\phantom{\rule{0.25em}{0ex}}A\mathrm{:\; 0}+0=0$

(b) 0.511 MeV

(c) The two $\gamma $ rays must travel in exactly opposite directions in order to conserve momentum, since initially there is zero momentum if the center of mass is initially at rest.

(a) ${}_{\text{88}}^{\text{226}}{\text{Ra}}_{138}\to {}_{\text{86}}^{\text{222}}{\text{Rn}}_{\text{136}}+{}_{2}^{4}{\text{He}}_{2}$

(b) 4.87 MeV

(a) 4.274 MeV

(b) $1\text{.}\text{927}\times {\text{10}}^{-5}$

(c) Since U-238 is a slowly decaying substance, only a very small number of nuclei decay on human timescales; therefore, although those nuclei that decay lose a noticeable fraction of their mass, the change in the total mass of the sample is not detectable for a macroscopic sample.

(a) ${}_{8}^{\text{15}}{\mathrm{O}}_{7}+{e}^{-}\to {}_{7}^{\text{15}}{\mathrm{N}}_{8}+{\nu}_{e}$

(b) 2.754 MeV

(a) $\text{1.23}\times {\text{10}}^{-3}$

(b) Only part of the emitted radiation goes in the direction of the detector. Only a fraction of that causes a response in the detector. Some of the emitted radiation (mostly $\alpha $ particles) is observed within the source. Some is absorbed within the source, some is absorbed by the detector, and some does not penetrate the detector.

(a) $1.68\times {10}^{\u20135}\phantom{\rule{0.25em}{0ex}}\text{Ci}$

(b) $8.65\times {10}^{10}\phantom{\rule{0.25em}{0ex}}\text{J}$

(c) $\$2.9\times {10}^{3}$

(a) 84.5 Ci

(b) An extremely large activity, many orders of magnitude greater than permitted for home use.

(c) The assumption of $1.00\phantom{\rule{0.25em}{0ex}}\text{\mu A}$ is unreasonably large. Other methods can detect much smaller decay rates.

(a) 7.680 MeV, consistent with graph

(b) 7.520 MeV, consistent with graph. Not significantly different from value for ${}^{\text{12}}\text{C}$, but sufficiently lower to allow decay into another nuclide that is more tightly bound.

(a) $1\text{.}\text{46}\times {\text{10}}^{-8}\phantom{\rule{0.25em}{0ex}}\mathrm{u}$ vs. 1.007825 u for ${}^{1}\text{H}$

(b) 0.000549 u

(c) $2\text{.}\text{66}\times {\text{10}}^{-5}$

(a) $\mathrm{\u20139.315\; MeV}$

(b) The negative binding energy implies an unbound system.

(c) This assumption that it is two bound neutrons is incorrect.

(a) ${}_{92}^{235}{\text{U}}_{143}\to {}_{\text{90}}^{\text{231}}{\text{Th}}_{\text{141}}+{}_{2}^{4}{\text{He}}_{\text{2}}$

(b) 4.679 MeV

(c) 4.599 MeV

a) $2.4\times {\text{10}}^{8}$ u

(b) The greatest known atomic masses are about 260. This result found in (a) is extremely large.

(c) The assumed radius is much too large to be reasonable.

(a) $\mathrm{\u20131.805\; MeV}$

(b) Negative energy implies energy input is necessary and the reaction cannot be spontaneous.

(c) Although all conversation laws are obeyed, energy must be supplied, so the assumption of spontaneous decay is incorrect.

(a) $\begin{array}{l}r={r}_{0}{A}^{1/3}=1.2{(235)}^{1/3}=7.4\text{fm}\\ \text{}V=\frac{4\pi {r}^{3}}{3}=1700{\text{fm}}^{3}\\ \text{}\rho =\frac{m}{v}=0.14{\text{u/fm}}^{3}\end{array}$

(b) For barium: $r={r}_{0}{A}^{1/3}=1.2{(142)}^{1/3}=6.3\text{fm}$

$\begin{array}{l}V=\frac{4\pi {r}^{3}}{3}=1047{\text{fm}}^{3}\\ \rho =\frac{m}{v}=0.14{\text{u/fm}}^{3}\end{array}$

For krypton: $r={r}_{0}{A}^{1/3}=1.2{(92)}^{1/3}=5.4\text{fm}$

$\begin{array}{l}V=\frac{4\pi {r}^{3}}{3}=660{\text{fm}}^{3}\\ \rho =\frac{m}{v}=0.14{\text{u/fm}}^{3}\end{array}$

To two significant figures, they are all alike.

## Test Prep for AP® Courses

When ${}_{95}^{241}\text{Am}$ undergoes α decay, it loses 2 neutrons and 2 protons. The resulting nucleus is therefore ${}_{93}^{237}\text{Np}$.

During this process, the nucleus emits a particle with -1 charge. In order for the overall charge of the system to remain constant, the charge of the nucleus must therefore increase by +1.

- No. Nucleon number is conserved (238 = 234 + 4), but the atomic number or charge is NOT conserved (92 ≠ 88+2).
- Yes. Nucleon number is conserved (223 = 209 + 14), and atomic number is conserved (88 = 82 + 6).
- Yes. Nucleon number is conserved (14 = 14), and charge is conserved if the electron’s charge is properly counted (6 = 7 + (-1)).
- No. Nucleon number is not conserved (24 ≠ 23). The positron released counts as a charge to conserve charge, but it doesn’t count as a nucleon.

This must be alpha decay since 4 nucleons (2 positive charges) are lost from the parent nucleus. The number remaining is found from:

$$\text{N}\left(t\right)={\text{N}}_{0}e\left(\frac{-0.693t}{\frac{{t}_{1}}{2}}\right)=3.4\times {10}^{17}e\left(\frac{-\left(0.693\right)\left(0.035\right)}{0.00173}\right)$$

$\text{N}\left(t\right)=4.1\times {10}^{11}$ nuclei