Question

A photon of 10.2 eV energy collides with a hydrogen atom in ground state inelastically. After few microseconds one more photon of energy 15 eV collides with the same hydrogen atom. Then what can be detected by a suitable detector.

• A. one photon of 10.2 eV and an electron of energy 1.4 eV
• B. 2 photons of energy 10.2 eV
• C. 2 photons of energy 3.4 eV
• D. 1 photon of 3.4 eV and one electron of 1.4 eV

Answer 1

Answer: (A)

one photon of 10.2 eV and an electron of energy 1.4 eV

Answer 2

• First Collision: A photon of 10.2 eV energy collides with a hydrogen atom in the ground state ($n=1$). The energy difference between the ground state ($E_1 = -13.6$ eV) and the first excited state ($E_2 = -3.4$ eV) is $E_2 - E_1 = -3.4 - (-13.6) = 10.2$ eV. Since the photon energy exactly matches this energy difference, the hydrogen atom absorbs the photon and gets excited to the $n=2$ state.

• Time Delay and De-excitation: The problem states that "After few microseconds" the second photon collides. The typical lifetime of an electron in the $n=2$ excited state of hydrogen is on the order of nanoseconds (around $1.6 \times 10^{-9}$ s). A few microseconds ($10^{-6}$ s) is much longer than this lifetime. Therefore, by the time the second photon arrives, the excited hydrogen atom will have already de-excited back to its ground state ($n=1$). During this de-excitation, it emits a photon with energy equal to the energy difference between $n=2$ and $n=1$, which is 10.2 eV. This photon can be detected.

• Second Collision: The hydrogen atom is now back in its ground state ($n=1$). A second photon of 15 eV energy collides with it. The ionization energy of a hydrogen atom from its ground state is 13.6 eV ($0 - E_1 = 0 - (-13.6) = 13.6$ eV). Since the incident photon energy (15 eV) is greater than the ionization energy (13.6 eV), the atom will be ionized. The excess energy will be given to the ejected electron as kinetic energy.
  Kinetic energy of the ejected electron = Photon energy - Ionization energy
  Kinetic energy of the ejected electron = 15 eV - 13.6 eV = 1.4 eV.
  This electron with 1.4 eV kinetic energy can be detected.

• Conclusion: A suitable detector would detect one photon of 10.2 eV (from the de-excitation after the first collision) and an electron of 1.4 eV (from the ionization after the second collision).