Option 1: Bohr atom explained through orbit relations and photon wavelengths
Step 1: Bohr assumed that the hydrogen electron can occupy only discrete non-radiating circular orbits (stationary states). The Coulomb pull equals the centripetal requirement, \( \frac{1}{4\pi\varepsilon_0}\frac{e^{2}}{r^{2}}=\frac{mv^{2}}{r} \), which fixes the speed on each orbit.
Step 2: He further restricted the orbits by quantising angular momentum, \(mvr = n h/2\pi\). Combining this with Step 1 gives quantised radii \(r_n \propto n^2\) and quantised energies \(E_n = -13.6/n^2\) eV, so the levels are \(E_1=-13.6\), \(E_2=-3.4\), \(E_3=-1.51\) eV.
Step 3: Radiation appears only during a jump, and the photon carries exactly the energy gap, \(h\nu = |E_i - E_f|\), equivalently \(\frac{1}{\lambda}=R\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)\). This third postulate is what turns discrete levels into sharp spectral lines.
Step 4 (Counting the lines): Working between the three levels n = 1, 2, 3, the number of distinct photon energies equals the number of level pairs, \(\binom{3}{2}=3\). So the emission spectrum shows three lines produced by the downward jumps \(3\to2\) (energy \(1.89\) eV, visible), \(3\to1\) (\(12.09\) eV, UV) and \(2\to1\) (\(10.2\) eV, UV).
Step 5: A cold hydrogen gas sits in the ground state n = 1, so it can only soak up photons that lift it upward. The two allowed upward jumps \(1\to3\) and \(1\to2\) remove exactly those two photon energies from a white beam, leaving two dark absorption lines. Thus emission = 3 lines, absorption = 2 lines.
Step 6 (Sketch): Ladder of levels with n = 1 at the bottom; downward arrows for the three emission lines and upward arrows from n = 1 for the two absorption lines.
Option 2: Neutron decay energy using the atomic-mass-unit method
Step 1: Einstein\'s relation \(E=mc^2\) says any loss of rest mass in a reaction is liberated as kinetic energy of the products; the total mass-energy before and after the decay must balance. Here the neutron is slightly heavier than proton plus electron, and that surplus mass is released.
Step 2: Treat the antineutrino as massless. The lost mass is \(\Delta m = m_n - m_p - m_e\). Plugging the data, \(\Delta m = 1.6747\times10^{-27} - 1.6725\times10^{-27} - 9.1\times10^{-31}\).
Step 3: \(\Delta m = 2.2\times10^{-30} - 0.91\times10^{-30} = 1.29\times10^{-30}\) kg.
Step 4 (Switch to amu): Using \(1\ \text{u}=1.66\times10^{-27}\) kg, \(\Delta m = \dfrac{1.29\times10^{-30}}{1.66\times10^{-27}} = 7.77\times10^{-4}\) u.
Step 5 (Use 1 u = 931 MeV): Energy released \(E = \Delta m \times 931\ \text{MeV} = 7.77\times10^{-4}\times931\).
Step 6: \(E \approx 0.72\ \text{MeV}\), which agrees with the \(E=\Delta m c^2\) value of \(0.726\) MeV found by the direct method.
\[\boxed{E \approx 0.72\ \text{MeV}}\]