Question

Enunciate the postulates of Bohr\'s model for the hydrogen atom. Show the spectral lines in the (i) emission and (ii) the absorption spectrum corresponding to the transition of hydrogen atom from \(n = 1\) to \(n = 3\) energy states.
OR
What is mass-energy conservation law? A neutron disintegrates into a proton, a \(\beta\)-particle and an antineutrino. The disintegration process is: \[ {}_{0}n^{1} \rightarrow {}_{1}H^{1} + {}_{-1}\beta^{0} + \bar{\nu} \] Find out the energy produced in this process in MeV. (Mass of neutron \(= 1.6747\times10^{-27}\) kg, Mass of proton \(= 1.6725\times10^{-27}\) kg, Mass of electron \(= 9.1\times10^{-31}\) kg.)

Hint:

Use Bohr\'s three postulates - non-radiating orbits, \(mvr=nh/2\pi\), and \(h\nu=E_i-E_f\) - and count 3 emission vs 2 absorption lines. For the OR part use \(E=\Delta m\,c^{2}\) with \(\Delta m = m_n-(m_p+m_e)\).

Answer 1

Option 1: Bohr postulates and the emission/absorption spectrum (n = 1 to n = 3)

Step 1 (Postulate of stationary orbits): In a hydrogen atom the electron revolves round the nucleus only in certain permitted circular orbits, called stationary orbits, without radiating energy. The electrostatic (Coulomb) attraction supplies the centripetal force: \[ \frac{1}{4\pi\varepsilon_0}\frac{e^{2}}{r^{2}} = \frac{mv^{2}}{r} \]Step 2 (Postulate of quantised angular momentum): Only those orbits are allowed for which the angular momentum of the electron is an integral multiple of \(h/2\pi\): \[ mvr = \frac{nh}{2\pi}, \qquad n = 1,2,3,\dots \]Step 3 (Frequency / energy condition): The atom emits or absorbs energy only when the electron jumps from one stationary orbit of energy \(E_i\) to another of energy \(E_f\). The photon frequency is given by \[ h\nu = E_i - E_f \]Step 4 (Energy levels): The energy of the \(n\)th level is \( E_n = -\dfrac{13.6}{n^{2}}\ \text{eV} \). Hence \(E_1=-13.6\ \text{eV}\), \(E_2=-3.4\ \text{eV}\), \(E_3=-1.51\ \text{eV}\).
Step 5 (Emission spectrum): In emission the electron falls from a higher level to a lower one, so within the levels n = 1, 2, 3 there are three downward transitions and hence three bright emission lines: \(3\rightarrow2\), \(2\rightarrow1\) and \(3\rightarrow1\). The lines ending on n = 1 (Lyman series) lie in the ultraviolet, while \(3\rightarrow2\) (Balmer, \(H_\alpha\)) lies in the visible region.
Step 6 (Absorption spectrum): In absorption the atom is normally in the ground state (n = 1) and absorbs photons to jump up. From n = 1 only two upward jumps are possible, \(1\rightarrow2\) and \(1\rightarrow3\), so two dark absorption lines appear on the bright continuous background.
Step 7 (Diagram): Draw three horizontal energy levels (n = 1 lowest, then n = 2, then n = 3). For emission draw three downward arrows \(3\rightarrow2,\ 3\rightarrow1,\ 2\rightarrow1\); for absorption draw two upward arrows \(1\rightarrow2,\ 1\rightarrow3\). Emission gives three lines, absorption gives two lines.

Option 2: Mass-energy conservation and energy released in neutron decay

Step 1 (Law): The mass-energy conservation law (Einstein) states that mass and energy are two forms of the same physical quantity and are interconvertible; the total mass-energy of an isolated system stays constant. A decrease in mass \(\Delta m\) appears as released energy through \( E = \Delta m\,c^{2} \).
Step 2 (Mass defect - formula): The antineutrino is (practically) massless, so the mass lost is \[ \Delta m = m_n - (m_p + m_e) \]Step 3 (Substitution): With \(m_e = 9.1\times10^{-31}=0.00091\times10^{-27}\) kg, \[ \Delta m = 1.6747\times10^{-27} - \left(1.6725\times10^{-27} + 0.00091\times10^{-27}\right) \]Step 4 (Arithmetic): \( \Delta m = (1.6747 - 1.67341)\times10^{-27} = 0.00129\times10^{-27} = 1.29\times10^{-30}\ \text{kg}. \)
Step 5 (Energy - substitution): \( E = \Delta m\,c^{2} = 1.29\times10^{-30}\times(3\times10^{8})^{2} = 1.29\times10^{-30}\times9\times10^{16}. \)
Step 6 (Energy in joule): \( E = 1.161\times10^{-13}\ \text{J}. \)
Step 7 (Convert to MeV): Since \(1\ \text{MeV}=1.6\times10^{-13}\ \text{J}\), \[ E = \frac{1.161\times10^{-13}}{1.6\times10^{-13}} = 0.726\ \text{MeV}. \]\[\boxed{E \approx 0.726\ \text{MeV}}\]