Two energy levels of a certain atom have energies \(-6.20\ \mathrm{eV}\) and \(-1.24\ \mathrm{eV}\). The atom makes a transition from the higher level to the lower level. What is the wavelength of the photon emitted? (Take \(hc=1240\ \mathrm{eV\,nm}\).)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(250\ \mathrm{nm}\)
When the atom falls to a lower level, it gives out one photon. The energy of this photon is the difference between the energies of the two levels. Its wavelength then follows from \( E=hc/\lambda \).
Given:
\( E_i=-1.24\ \mathrm{eV} \)
\( E_f=-6.20\ \mathrm{eV} \)
Energy of the photon:
\( \begin{aligned} E &= E_i-E_f \\ &= -1.24-(-6.20) \\ &= 4.96\ \mathrm{eV} \end{aligned} \)
Wavelength of the photon:
\( \begin{aligned} \lambda &= \frac{hc}{E} \\ &= \frac{1240}{4.96} \\ &= 250\ \mathrm{nm} \end{aligned} \)
From the above, the photon carries \( 4.96\ \mathrm{eV} \). Hence, its wavelength is \( 250\ \mathrm{nm} \).
NCERT (2026-27):
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”
\( \displaystyle h\nu=E_i-E_f \)\( \qquad (12.6) \)