Atoms and Nuclei Questions and Answers

Answer: A

The transition gives the shortest wavelength of the Paschen series.

The Paschen series contains the transitions ending on \(n_{1} = 3\). The photon energy is largest, and so the wavelength shortest, when the electron falls from the highest level, \(n_{2} = \infty\) (the series limit):

\[ \begin{aligned} \frac{1}{\lambda} \\ &= R\left(\frac{1}{3^{2}} - \frac{1}{\infty^{2}}\right) \\ &= \frac{R}{9} \end{aligned} \]

The longest Paschen wavelength comes from \(4 \to 3\). The Balmer series ends on \(n = 2\) and the Brackett series on \(n = 4\), so they do not contain this transition.

Answer: D

The transition is \(n_{2} = \infty\) and \(n_{1} = 4\).

The Brackett series consists of transitions ending on n = 4. Its wavelengths satisfy \(\frac{1}{\lambda} = R\left(\frac{1}{4^{2}} - \frac{1}{n_{2}^{2}}\right)\). The shortest wavelength means the largest energy gap, which is from \(n_{2} = \infty\), the series limit.

5 to 4 is the longest wavelength of the Brackett series. Transitions ending on n = 1, 2 or 3 belong to the Lyman, Balmer and Paschen series.

Answer: A

The half-life is 16 years.

Find what fraction remains:

\[ \begin{aligned} \frac{0.75 \times 10^{5}}{6 \times 10^{5}} &= \frac{1}{8} \\ &= \left(\frac{1}{2}\right)^{3} \end{aligned} \]

So three half-lives have passed in 48 years.

\[ T_{1/2} = \frac{48}{3} = 16\ \text{years} \]

24 years assumes two half-lives (a quarter left), 12 years four (a sixteenth left), and 6 years eight half-lives.

Answer: B

Z is a \(\beta^{+}\) particle (a positron).

The mass number stays 197 while the atomic number falls from 80 to 79. Balancing charge and mass number, Z has charge +1 and mass number 0: a positron. Positron emission is accompanied by a neutrino, the v in the equation.

An α-particle would lower the mass number by 4, \(\beta^{-}\) would raise Z by 1, a proton would lower the mass number by 1, and a neutron would lower the mass number without changing Z.

Answer: D

The fraction remaining is \(\frac{1}{e}\).

Radioactive decay follows \(N = N_{0}e^{-\lambda t}\), and the mean life is \(\tau = \frac{1}{\lambda}\). After a time \(t = \tau\):

\[ \frac{N}{N_{0}} = e^{-1} = \frac{1}{e} \approx 0.37 \]

\(1 - \frac{1}{e}\) is the fraction that has decayed, and 1 − e and 2e are not fractions between 0 and 1.

Answer: A

The transition is \(n = 3\) to \(n = 2\).

By the Rydberg formula, \(\frac{1}{\lambda} = R\left(\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}\right)\). Here \(\frac{1}{\lambda} = \frac{5R}{36}\). For \(n_{2} = 3\) to \(n_{1} = 2\):

\(R\left(\frac{1}{4} - \frac{1}{9}\right) = \frac{5R}{36}\)

This matches, and it is the first Balmer line, H-alpha. 4 to 2 gives \(\frac{3R}{16}\), 3 to 1 gives \(\frac{8R}{9}\), 4 to 1 gives \(\frac{15R}{16}\), and 4 to 3 gives \(\frac{7R}{144}\); none equals \(\frac{5R}{36}\).

Answer: B

The half-life is \(\frac{t}{2}\) years.

When 75% has decayed, 25% remains, which is \(\frac{1}{4} = \left(\frac{1}{2}\right)^{2}\). So two half-lives have passed in t years:

\[ 2T_{1/2} = t \Rightarrow T_{1/2} = \frac{t}{2} \]

The trap is to read 75% decayed as 75% remaining, or to take t itself as the half-life, which would leave 50% undecayed. \(\frac{t}{4}\) assumes four half-lives.

Answer: A

The angular momentum is \(\frac{h}{\pi}\).

Bohr's quantisation condition gives \(L = \frac{nh}{2\pi}\). For the second orbit, \(n = 2\):

\[ L = \frac{2h}{2\pi} = \frac{h}{\pi} \]

\(\frac{h}{2\pi}\) is the first orbit, \(\frac{3h}{2\pi}\) the third, and \(\frac{2h}{\pi}\) would be the fourth.

In general the angular momentum in the \(n\)th orbit is an integral multiple of \(\frac{h}{2\pi}\), so successive orbits differ by exactly \(\frac{h}{2\pi}\).

Answer: A

X is a neutrino.

This is \(\beta^{+}\) decay: a proton in the nucleus becomes a neutron (Z falls from 11 to 10, A stays 22), emitting a positron and a neutrino, \(p \rightarrow n + e^{+} + \nu\).

The antineutrino accompanies \(\beta^{-}\) decay (electron emission). Charge and nucleon number already balance with the positron, so X cannot be a further electron, positron or neutron.

Checking conservation: mass number \(22 = 22 + 0\) and charge \(11 = 10 + 1\), so X has zero charge and zero mass number, as a neutrino does.

Answer: C

The ratio is 5:27.

The longest wavelength in each series comes from the smallest energy jump into its lowest level.

\[ \begin{aligned} \text{Lyman } (2 \to 1): \frac{1}{\lambda_{L}} &= R\left(1 - \tfrac{1}{4}\right) = \tfrac{3R}{4} \\ \text{Balmer } (3 \to 2): \frac{1}{\lambda_{B}} &= R\left(\tfrac{1}{4} - \tfrac{1}{9}\right) = \tfrac{5R}{36} \end{aligned} \]

\[ \begin{aligned} \frac{\lambda_{L}}{\lambda_{B}} &= \frac{5R/36}{3R/4} \\ &= \frac{5}{27} \end{aligned} \]

Inverting the ratio by mistake gives 27 : 5. The options with 29 do not come from the Rydberg formula.