In a hydrogen atom, which of the following electronic transitions would involve the maximum energy change?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → From n = 3 to n = 1
By Bohr’s third postulate, the energy change in a transition is the difference between the energies of the initial and final levels, \( \Delta E=E_i-E_f \). The levels of hydrogen are \( E_n=-13.6\,\mathrm{eV}/n^{2} \), so they lie far apart near n = 1 and crowd together higher up.
The levels involved are:
\( E_1=-13.6\,\mathrm{eV} \)
\( E_2=-3.40\,\mathrm{eV} \)
\( E_3\approx-1.51\,\mathrm{eV} \)
\( E_4=-0.85\,\mathrm{eV} \)
The energy change in each transition is:
\( \begin{aligned} E_2-E_1 &= -3.40\,\mathrm{eV}-(-13.6\,\mathrm{eV}) \\ &= 10.2\,\mathrm{eV} \end{aligned} \)
\( \begin{aligned} E_3-E_1 &= -1.51\,\mathrm{eV}-(-13.6\,\mathrm{eV}) \\ &\approx 12.09\,\mathrm{eV} \end{aligned} \)
\( \begin{aligned} E_4-E_2 &= -0.85\,\mathrm{eV}-(-3.40\,\mathrm{eV}) \\ &= 2.55\,\mathrm{eV} \end{aligned} \)
\( \begin{aligned} E_3-E_2 &= -1.51\,\mathrm{eV}-(-3.40\,\mathrm{eV}) \\ &\approx 1.89\,\mathrm{eV} \end{aligned} \)
From the above, the jump from n = 3 to n = 1 has the largest energy change, about 12.09 eV. It ends on the ground state, where the levels are farthest apart. Hence, the maximum energy change is in the transition from n = 3 to n = 1.
NCERT (2026-27):
Bohr Model of the Hydrogen Atom – “(iii) Bohr’s third postulate incorporated into atomic theory the early quantum concepts that had been developed by Planck and Einstein. It states that an electron might make a transition from one of its specified non-radiating orbits to another of lower energy. When it does so, a photon is emitted having energy equal to the energy difference between the initial and final states. The frequency of the emitted photon is then given by
\( h\nu=E_i-E_f \qquad (12.6) \)
where Ei and Ef are the energies of the initial and final states and Ei > Ef.”
Bohr Model of the Hydrogen Atom – “Atomic energies are often expressed in electron volts (eV) rather than joules. Since 1 eV = 1.6 × 10–19 J, … can be rewritten as
\( \displaystyle E_n=-\frac{13.6\,\mathrm{eV}}{n^{2}} \)\( \qquad (12.10) \)”