Nuclear Properties and Structure Questions and Answers

Answer: B

The minimum photon energy is \(B\left(1 + \dfrac{B}{2mc^{2}}\right)\).

The photon must supply the binding energy B and also the recoil kinetic energy of the products, because its momentum E/c must be conserved. At threshold the two nucleons move off together with total momentum E/c and total rest energy mc² + B. Energy conservation with the bound state at rest:

\[ \begin{aligned} E + mc^{2} &= \sqrt{(mc^{2} + B)^{2} + E^{2}} \\ E &= B + \frac{B^{2}}{2mc^{2}} = B\left(1 + \frac{B}{2mc^{2}}\right) \end{aligned} \]

B alone ignores the recoil. The minus sign would make the threshold smaller than the binding energy, which energy conservation forbids, and B - mc² is negative for any real nucleus.

Answer: B

The ratio of nucleon numbers is 8.

Nuclear radius follows \( R = r_{0}A^{1/3} \), so \( A \propto R^{3} \):

\[ \begin{aligned} \frac{A_{A}}{A_{Al}} &= \left(\frac{R_{A}}{R_{Al}}\right)^{3} \\ &= 2^{3} = 8 \end{aligned} \]

Nucleus A has 8 × 27 = 216 nucleons. 16 would assume \( A \propto R^{4} \), and 40 or 14 do not come from the cube law.

Answer: B

The wavelength is nearly 9 F.

A 10 MeV proton is non-relativistic (its rest energy is 938 MeV), so its momentum is \(p = \sqrt{2mK}\):

\[ pc = \sqrt{2 \times 938 \times 10} \approx 137\ \text{MeV} \]

\[ \begin{aligned} \lambda &= \frac{hc}{pc} = \frac{1240\ \text{MeV fm}}{137\ \text{MeV}} \\ &\approx 9.0\ \text{fm} \end{aligned} \]

Since \(\lambda \propto 1/\sqrt{K}\), the other values belong to other energies: 7 F to about 17 MeV, 11 F to about 7 MeV and 13 F to about 5 MeV.

Answer: B

The binding energy per nucleon is nearly 8.54 MeV/nucleon.

Silver-107 has Z = 47 protons and N = 107 - 47 = 60 neutrons. The mass defect, using the hydrogen-atom mass for the proton so that the electrons cancel:

\[ \begin{aligned} \Delta m &= 47(1.007825) + 60(1.008665) \\ &\quad - 106.905091 \\ &= 107.887675 - 106.905091 \\ &= 0.982584\ \text{amu} \\ B &= 0.982584 \times 931.5 = 915.3\ \text{MeV} \\ B/A &= 915.3/107 = 8.55\ \text{MeV} \end{aligned} \]

That is close to the 8.5 MeV typical of medium nuclei. 8.10, 8.95 and 9.25 MeV/nucleon do not follow from these masses.

Answer: A

The binding energy per nucleon is about 8.7 MeV.

\({}^{55}_{25}Mn\) has 25 protons and 30 neutrons. The mass defect is

\[ \begin{aligned} \Delta m &= 25(1.007825) + 30(1.008665) - 54.938050 \\ &= 25.195625 + 30.259950 - 54.938050 \\ &= 0.517525\ \text{amu} \end{aligned} \]

\[ \begin{aligned} B &= 0.517525 \times 931.5 = 482.1\ \text{MeV} \\ B/A &= 482.1/55 = 8.77\ \text{MeV} \end{aligned} \]

which is the 8.7 MeV option. The other values exceed the highest binding energy per nucleon of any nucleus, about 8.8 MeV near iron and nickel, so they cannot be right for manganese.

Answer: B

On the key, the electron needs an energy of about 0.2 GeV.

To resolve structure of size R, the probe's reduced de Broglie wavelength must be no larger than R, so \(p \approx \hbar/R\). At these energies the electron is ultra-relativistic and E = pc:

\[ \begin{aligned} E &\approx \frac{\hbar c}{R} = \frac{197\ \text{MeV fm}}{1\ \text{fm}} \\ &\approx 0.2\ \text{GeV} \end{aligned} \]

using \(R = 10^{-13}\) cm = 1 fm. Using the full wavelength \(\lambda = h/p\) instead would give about 1.2 GeV, which is not offered; the key's estimate is the order of magnitude used in electron-scattering experiments such as Hofstadter's. 0.1 GeV gives too long a wavelength, and 0.3 or 0.4 GeV are not what the estimate yields.

Answer: C

The binding energy is closest to 556.18 MeV.

The nucleus has 29 protons and 35 neutrons. The mass defect, using the given masses, is:

\[ \begin{aligned} \Delta m &= 29(1.007825) + 35(1.008665) - 63.9298 \\ &= 29.226925 + 35.303275 - 63.9298 \\ &= 0.6004\ \text{u} \end{aligned} \]

Multiplying by 931.5 MeV per u gives about 559 MeV, about 8.7 MeV per nucleon, typical of nuclei near iron. Of the options, 556.18 MeV is the only one within 1%; 450, 512 and 612 MeV are far from the mass defect.

Answer: D

The ratio is 0.

The pairing term depends on whether the proton and neutron numbers are even or odd:

• odd \(A\): \(\delta = 0\);
• even \(Z\), even \(N\): \(\delta = +a_{p}A^{-1/2}\) (or a similar power of \(A\));
• odd \(Z\), odd \(N\): \(\delta = -a_{p}A^{-1/2}\).

\({}^{15}_{7}\mathrm{N}\) has odd mass number (\(Z = 7\), \(N = 8\)), so its pairing term is zero. \({}^{256}_{100}\mathrm{Fm}\) has \(Z = 100\) and \(N = 156\), both even, so its pairing term is non-zero. The ratio is therefore 0.

Non-zero values such as \(\pm 4\) or \(\pm 8\) could only arise if both nuclei had non-zero pairing terms, comparing their \(A^{-1/2}\) factors, which is not the case here.

Answer: A

The charge density is \(\rho_{ch}(r) = \frac{\rho_{0}}{1 + \exp\left(\frac{r - R}{a}\right)}\).

This is the Fermi (Woods-Saxon) distribution fitted to electron-scattering data. Check its shape:

• deep inside (\(r \ll R\)) the exponential is tiny, so \(\rho \approx \rho_{0}\), a constant core density;
• at \(r = R\) the density is \(\rho_{0}/2\), so \(R\) is the half-density radius;
• far outside (\(r \gg R\)) the exponential grows and \(\rho\) falls smoothly to zero over a skin set by \(a\).

With \(r + R\) in the exponent the density would never reach \(\rho_{0}\) and would have no surface at \(R\). With \(1 - \exp\) in the denominator it would blow up at \(r = R\) and turn negative outside, which a density cannot do.