The electron of a hydrogen atom is in the first excited state. It absorbs a photon, and its angular momentum becomes \(3/2\) times its earlier value. The energy of the photon is (Take \(E_n=-13.6/n^2\ \mathrm{eV}\))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(1.89\ \mathrm{eV}\)
The first excited state is \( n=2 \). Bohr’s rule \( L=nh/2\pi \) shows that the angular momentum is proportional to \( n \). The photon absorbed must supply the energy difference of the two states.
\( \begin{aligned} \frac{L_f}{L_i} &= \frac{n_f}{2} \\ &= \frac{3}{2} \end{aligned} \)
\( n_f=3 \)
\( \begin{aligned} E_2 &= -\frac{13.6}{2^2} \\ &= -3.40\ \mathrm{eV} \end{aligned} \)
\( \begin{aligned} E_3 &= -\frac{13.6}{3^2} \\ &= -1.51\ \mathrm{eV} \end{aligned} \)
\( \begin{aligned} h\nu &= E_3-E_2 \\ &= -1.51-(-3.40) \\ &= 1.89\ \mathrm{eV} \end{aligned} \)
From the above, the photon energy is \( 1.89\ \mathrm{eV} \).
NCERT (2026-27):
Bohr Model of the Hydrogen Atom – “… This postulate states that the electron revolves around the nucleus only in those orbits for which the angular momentum is some integral multiple of h/2π where h is the Planck’s constant (= 6.6 × 10–34 J s). Thus the angular momentum (L) of the orbiting electron is quantised. That is”
\( \displaystyle L=\frac{nh}{2\pi} \)\( \qquad (12.5) \)
Summary – “An atom absorbs radiation of the same frequency the atom emits, in which case the electron is transferred to an orbit with a higher value of n.”
\( E_i+h\nu=E_f \)