Atoms and Nuclei Questions and Answers

Answer: E

The wrong statement is Model is applicable for many electron systems also.

Bohr's model works only for hydrogen and hydrogen-like ions with a single electron, such as He⁺ and Li²⁺. It cannot account for the spectra of atoms with more than one electron, because it ignores electron-electron repulsion.

The other statements describe the model correctly: electrons move in circular orbits, those stationary orbits do not radiate energy, angular momentum is quantised as nh/2π, and the model applies only to one-electron systems.

Answer: C

The isotones are a, c and d.

Isotones have the same number of neutrons, \(N = A - Z\):

\[ \begin{aligned} {}_{6}^{14}C &: 14 - 6 = 8 \\ {}_{7}^{14}N &: 14 - 7 = 7 \\ {}_{7}^{15}N &: 15 - 7 = 8 \\ {}_{8}^{16}O &: 16 - 8 = 8 \end{aligned} \]

Carbon-14, nitrogen-15 and oxygen-16 all have 8 neutrons. Nitrogen-14 has 7, so any option containing b is wrong; b and c are isotopes (same Z), and a and b are isobars (same A).

Answer: B

The speed in the third orbit is \(\dfrac{v}{3}\).

In Bohr's model the electron's speed in the nth orbit is \(v_{n} = \frac{e^{2}}{2\varepsilon_{0}hn}\), proportional to \(\frac{1}{n}\) for hydrogen (about \(2.2 \times 10^{6}\ \text{m s}^{-1}\) divided by \(n\)). So the third orbit has one third the speed of the first.

\(\frac{v}{9}\) confuses speed with energy, which goes as \(\frac{1}{n^{2}}\), and \(3v\) inverts the dependence: outer orbits are slower, not faster.

Answer: C

\(a\) and \(b\) are 228 and 92.

Each alpha emission lowers the mass number by 4 and the atomic number by 2; each beta-minus emission leaves the mass number unchanged and raises the atomic number by 1.

\[ \begin{aligned} a &= 236 - 2 \times 4 = 228 \\ b &= 94 - 2 \times 2 + 2 \times 1 = 92 \end{aligned} \]

Only the alpha decays change the mass number, so any option other than 228 has the mass number wrong; 226 and 92 and 230 and 92 get the atomic number right but not the mass number.

Answer: B

The difference is \(\frac{h}{\pi}\).

Bohr's quantisation condition is \(L = \frac{nh}{2\pi}\). So

\[ \begin{aligned} L_{5} - L_{3} &= \frac{5h}{2\pi} - \frac{3h}{2\pi} \\ &= \frac{2h}{2\pi} = \frac{h}{\pi} \end{aligned} \]

\(\frac{5h}{2\pi}\) and \(\frac{3h}{2\pi}\) are the two angular momenta themselves, and \(\frac{2h}{\pi}\) is twice the difference.

Answer: C

The relations are \(r \propto n^{2}\) and \(v \propto \frac{1}{n}\).

In Bohr's model of hydrogen, \(r_n = 0.529\,n^{2}\) Å and \(v_n = \frac{2.18 \times 10^{6}}{n}\) m/s. They follow from \(mvr = \frac{nh}{2\pi}\) together with the condition that the Coulomb force supplies the centripetal force.

Higher orbits are larger and slower, so any option in which v increases with n is wrong, and r grows as the square of n, not linearly.

Answer: B

The half-life is 64 days.

6.25% is \(\frac{1}{16} = \left(\frac{1}{2}\right)^{4}\) of the original amount, so four half-lives have passed.

\[ T_{1/2} = \frac{256}{4} = 64\ \text{days} \]

With a half-life of 126 days about 25% would remain after 256 days, and half-lives of 50 or 25 days would leave far less than 6.25%. The general rule is \(N = N_0\left(\frac{1}{2}\right)^{t/T}\), so \(\frac{t}{T} = 4\) here.

Answer: A

A hydrogen bomb uses fusion.

In a thermonuclear (hydrogen) bomb, isotopes of hydrogen, deuterium and tritium, are fused into helium at extremely high temperature, releasing enormous energy. The temperature needed is produced by a fission bomb used as the trigger.

Fission, the splitting of heavy nuclei such as uranium-235, is the principle of the atom bomb. Ionisation and electrolysis are not nuclear processes at all.

Answer: C

The energy released per fission of uranium-235 is nearly 200 MeV.

When a U-235 nucleus absorbs a neutron and splits, the products together have less mass than the original nucleus and neutron. That mass defect, about 0.2 atomic mass units, appears as energy by \(E = mc^{2}\), and since 1 u corresponds to about 931 MeV, the release is roughly 200 MeV per fission.

200 eV and 2000 eV are chemical-scale energies, a million times too small, and 20 MeV is an order of magnitude short.

Answer: B

The energy released is about \(4 \times 10^{15}\) J.

Count the nuclei in 50 kg of uranium-235, then multiply by the energy per fission.

\[ \begin{aligned} N &= \frac{50000}{235} \times 6.02 \times 10^{23} \\ &\approx 1.28 \times 10^{26} \\ E &= 200 \times 1.6 \times 10^{-13}\ \text{J per fission} \\ &= 3.2 \times 10^{-11}\ \text{J} \\ NE &\approx 4.1 \times 10^{15}\ \text{J} \end{aligned} \]

\(4 \times 10^{12}\) J would be the answer for 50 g, and \(4 \times 10^{13}\) J for 500 g, so they come from using the wrong mass unit.