The electron in a hydrogen atom jumps from the \(n=5\) level to the \(n=2\) level. What is the wavelength of the photon emitted? (Take \(E_n=-13.6/n^2\ \mathrm{eV}\) and \(hc=1240\ \mathrm{eV\,nm}\).)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(434\ \mathrm{nm}\)
The energy of each level comes from \( E_n=-13.6/n^2 \) eV. The photon takes away the difference between the two levels, and its wavelength is \( \lambda=hc/E \).
Energies of the two levels:
\( \begin{aligned} E_5 &= -\frac{13.6}{5^2} \\ &= -0.544\ \mathrm{eV} \end{aligned} \)
\( \begin{aligned} E_2 &= -\frac{13.6}{2^2} \\ &= -3.40\ \mathrm{eV} \end{aligned} \)
Energy of the photon:
\( \begin{aligned} E &= E_5-E_2 \\ &= -0.544-(-3.40) \\ &= 2.86\ \mathrm{eV} \end{aligned} \)
Wavelength of the photon:
\( \begin{aligned} \lambda &= \frac{hc}{E} \\ &= \frac{1240}{2.86} \\ &= 434\ \mathrm{nm} \end{aligned} \)
Therefore, the jump from \( n=5 \) to \( n=2 \) gives a photon of wavelength about \( 434\ \mathrm{nm} \), in the visible region.
NCERT (2026-27):
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”
\( \displaystyle E_n=-\frac{13.6\ \mathrm{eV}}{n^{2}} \)\( \qquad (12.10) \)
Bohr Model of the Hydrogen Atom – “… 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”
\( \displaystyle h\nu=E_i-E_f \)\( \qquad (12.6) \)