A hydrogen atom is in its first excited state \((n=2)\). It absorbs a photon of wavelength \(410\ \mathrm{nm}\). The principal quantum number of the level to which the electron is raised is (Take \(E_n=-13.6/n^2\ \mathrm{eV}\) and \(hc=1240\ \mathrm{eV\,nm}\).)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 6
On absorbing a photon, the atom gains the photon energy. So the energy of the new level is the energy of the \( n=2 \) level plus the photon energy. The value of \( n \) then follows from \( E_n=-13.6/n^2 \) eV.
Energy of the photon:
\( \begin{aligned} E &= \frac{hc}{\lambda} \\ &= \frac{1240}{410} \\ &= 3.02\ \mathrm{eV} \end{aligned} \)
Energy of the new level:
\( \begin{aligned} E_f &= E_2+E \\ &= -3.40+3.02 \\ &= -0.38\ \mathrm{eV} \end{aligned} \)
Quantum number of the new level:
\( \begin{aligned} n^2 &= \frac{13.6}{0.38} \\ &\approx 36 \\ n &= 6 \end{aligned} \)
From the above, the electron reaches the level with energy \( -0.38\ \mathrm{eV} \). Hence, it is raised to \( n=6 \).
NCERT (2026-27):
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.”
\( \displaystyle E_i+h\nu=E_f \)
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”
\( \displaystyle E_n=-\frac{13.6\ \mathrm{eV}}{n^{2}} \)\( \qquad (12.10) \)