Atoms and Nuclei Questions and Answers

Answer: E

The binding energy is approximately 28.4 MeV.

Helium-4 has 2 protons and 2 neutrons. The mass defect is

\[ \begin{aligned} \Delta m &= 2(1.0073) + 2(1.0087) - 4.0015 \\ &= 4.0320 - 4.0015 = 0.0305\ u \\ E_{b} &= 0.0305 \times 931.5 \approx 28.4\ \text{MeV} \end{aligned} \]

This is about 7.1 MeV per nucleon, the well-known value for helium-4. The option 2.84 MeV slips a decimal place.

Answer: D

Nucleons are bound by short-range nuclear forces.

The strong nuclear force holds protons and neutrons together. It is the strongest of the fundamental forces but acts only over very short distances, about 2-3 femtometres, roughly the size of a nucleus, and it is nearly the same between any pair of nucleons whether charged or not.

Electromagnetic (electrostatic) forces push protons apart, so they work against binding. Gravity is far too weak at this scale. Nuclear forces are short-range, not long-range, which is why very large nuclei become unstable.

Answer: C

The values are \(\dfrac{h}{\pi}\) and -3.4 eV.

In Bohr's model the angular momentum is quantised, \(L = \frac{nh}{2\pi}\), and the energy is \(E_{n} = -\frac{13.6}{n^{2}}\) eV. For \(n = 2\):

\[ \begin{aligned} L &= \frac{2h}{2\pi} = \frac{h}{\pi} \\ E_{2} &= -\frac{13.6}{4} = -3.4\ \text{eV} \end{aligned} \]

-13.6 eV is the ground state (n = 1), and -1.5 eV is about the third orbit (n = 3). \(\frac{h}{2\pi}\) is the angular momentum of the first orbit, and \(\frac{2h}{\pi}\) that of the fourth.

Answer: C

The ratio is 27 : 5.

Wave number \(\bar{\nu} = R\left(\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}\right)\). The first Lyman line is 2 → 1, and the first Balmer line is 3 → 2.

\[ \begin{aligned} \text{Lyman} &: 1 - \tfrac{1}{4} = \tfrac{3}{4} \\ \text{Balmer} &: \tfrac{1}{4} - \tfrac{1}{9} = \tfrac{5}{36} \\ \tfrac{3}{4} : \tfrac{5}{36} &= 27 : 5 \end{aligned} \]

5 : 27 is the ratio of the wavelengths, since wavelength is the reciprocal of wave number.

Answer: D

It takes t = 5T.

When 1550 of the 1600 atoms have decayed, 50 remain.

\[ \begin{aligned} \frac{N}{N_{0}} &= \frac{50}{1600} = \frac{1}{32} \\ &= \left(\frac{1}{2}\right)^{5} \end{aligned} \]

So five half-lives have passed: 1600 → 800 → 400 → 200 → 100 → 50. 4T leaves 100 atoms and 6T leaves 25.

A quick count: each half-life halves the number left, so after n half-lives the fraction remaining is \(\frac{1}{2^{n}}\). Finding the n with \(2^{n} = \frac{1600}{50} = 32\) gives n = 5 directly.

Answer: E

The energy is -0.85 eV.

The ground state is \(n = 1\), so the first, second and third excited states are \(n = 2, 3, 4\). The energy levels of hydrogen are \(E_n = -\frac{13.6}{n^{2}}\) eV.

\[ E_4 = -\frac{13.6}{16} = -0.85 \text{ eV} \]

-1.51 eV is the n = 3 level, the second excited state, which is the trap of taking 'third excited' as n = 3. -3.4 eV is n = 2 and -13.6 eV the ground state.

Answer: B

The expected mass is 12.0956 u.

A carbon-12 nucleus contains 6 protons and 6 neutrons. Adding the masses of these separate nucleons gives the expected mass (the actual nuclear mass is smaller by the mass defect).

\[ \begin{aligned} m &= 6(1.00727) + 6(1.00866) \\ &= 6.04362 + 6.05196 \\ &= 12.09558 \approx 12.0956 \text{ u} \end{aligned} \]

6.0478 u counts only six nucleons, and 2.0159 u is one proton plus one neutron. The other values use the wrong number of particles.

Answer: D

The radius of the third orbit is 4.77 Å.

In the Bohr model, orbit radii grow as the square of the principal quantum number, \(r_n = n^{2}r_1\):

\[ r_3 = 9 \times 0.53 = 4.77\ \text{Å} \]

1.59 Å takes the radius as proportional to \(n\), 2.38 Å is not a multiple from either rule, and 9.54 Å doubles the right answer.

Answer: A

The energy released is about \(1.2 \times 10^{24}\) MeV.

2.35 g of ²³⁵U is \(\frac{2.35}{235} = 0.01\) mol, which contains

\[ \begin{aligned} N &= 0.01 \times 6.02 \times 10^{23} = 6.02 \times 10^{21} \\ E &= 6.02 \times 10^{21} \times 200 \\ &\approx 1.2 \times 10^{24}\ \text{MeV} \end{aligned} \]

\(0.6 \times 10^{24}\) MeV is the number of nuclei times 100 instead of 200, and \(2.4 \times 10^{24}\) MeV doubles the result.

Answer: C

The total binding energy is 490 MeV.

Binding energy per nucleon is the total binding energy divided by the mass number, so multiply back.

\[ BE = 8.75 \times 56 = 490\ \text{MeV} \]

56 MeV is just the mass number, 8 MeV is roughly the per-nucleon value, and 504 MeV would be 9 MeV per nucleon. A nucleus with A = 56 is iron-56, close to the peak of the binding energy curve.