An atom is in its lowest energy level A. It absorbs a photon of wavelength \(310\ \mathrm{nm}\) and goes up to a level C. It then falls from C to a level B and emits a photon of wavelength \(1240\ \mathrm{nm}\). By how much is the energy of level B higher than that of level A? (Take \(hc=1240\ \mathrm{eV\,nm}\).)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(3.0\ \mathrm{eV}\)
When the atom absorbs a photon, its energy rises by the photon energy. When it emits a photon, its energy falls by the photon energy. So we work with energies, and each photon energy is found from \( E=hc/\lambda \).
Photon absorbed (A to C):
\( \begin{aligned} E_C-E_A &= \frac{hc}{\lambda_1} \\ &= \frac{1240}{310} \\ &= 4.0\ \mathrm{eV} \end{aligned} \)
Photon emitted (C to B):
\( \begin{aligned} E_C-E_B &= \frac{hc}{\lambda_2} \\ &= \frac{1240}{1240} \\ &= 1.0\ \mathrm{eV} \end{aligned} \)
Energy of B above A:
\( \begin{aligned} E_B-E_A &= (E_C-E_A)-(E_C-E_B) \\ &= 4.0-1.0 \\ &= 3.0\ \mathrm{eV} \end{aligned} \)
From the above, the atom gains \( 4.0\ \mathrm{eV} \) and gives back only \( 1.0\ \mathrm{eV} \). Hence, level B lies \( 3.0\ \mathrm{eV} \) above level A.
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.”
\( \displaystyle E_i+h\nu=E_f \)
Summary – “(c) The third postulate 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 (ν) of the emitted photon is then given by”
\( \displaystyle h\nu=E_i-E_f \)