Modern Physics & Quantum

According to Bohr's model of the hydrogen atom, for an electron orbiting in a stationary orbit:

• Kinetic energy: $K = \frac{1}{8\pi\varepsilon_0} \frac{e^2}{r} \gt 0$
• Potential energy: $U = -\frac{1}{4\pi\varepsilon_0} \frac{e^2}{r} = -2K$
• Total mechanical energy: $E = K + U = K + (-2K) = -K$
  Given that kinetic energy is $K$, the potential energy is $-2K$ and the total energy is $-K$.

• The radius of a nucleus of mass number $A$ is given by $R = R_0 A^{1/3}$, where $R_0 \approx 1.2\,\text{fm}$.
• The volume of the nucleus is:
  $$V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (R_0 A^{1/3})^3 = \frac{4}{3}\pi R_0^3 A$$
• The mass of the nucleus is $M \approx A \cdot m_N$, where $m_N$ is the average mass of a nucleon.
• Therefore, the nuclear density is:
  $$\rho = \frac{M}{V} = \frac{A \cdot m_N}{\frac{4}{3}\pi R_0^3 A} = \frac{m_N}{\frac{4}{3}\pi R_0^3} = \text{constant} \approx 2.3 \times 10^{17}\,\text{kg/m}^3$$
  Since the nuclear density is completely independent of the mass number $A$, the ratio of densities of any two nuclei is $1 : 1$.

The shortest wavelength (series limit, $n_2 = \infty$) for a spectral series with lower level $n_1$ is given by Rydberg's formula:

$$\frac{1}{\lambda_{\min}} = R_H \left(\frac{1}{n_1^2} - \frac{1}{\infty^2}\right) = \frac{R_H}{n_1^2} \implies \lambda_{\min} = \frac{n_1^2}{R_H}$$

Calculating the shortest wavelength for each series:

• Lyman series ($n_1 = 1$): $\lambda_{\min} = \frac{1}{R_H} = \frac{1}{R_H}$
• Balmer series ($n_1 = 2$): $\lambda_{\min} = \frac{4}{R_H}$
• Brackett series ($n_1 = 4$): $\lambda_{\min} = \frac{16}{R_H}$
• Pfund series ($n_1 = 5$): $\lambda_{\min} = \frac{25}{R_H}$
  Arranging in decreasing order of wavelength (largest to smallest):
  $$\text{Pfund (A)} \gt \text{Brackett (C)} \gt \text{Balmer (B)} \gt \text{Lyman (D)}$$
  $$(A), (C), (B), (D)$$