Question

The recoil speed of a hydrogen atom after it goes from \( n = 5 \) state to \( n = 1 \) state will be:

• A. 4.34 m/s
• B. 2.19 m/s
• C. 4.17 m/s
• D. 3.25 m/s

Hint:

When a photon is emitted during an atomic transition, the recoil speed of the atom can be found using the conservation of momentum. The photon's energy and momentum are crucial to calculate the recoil velocity of the atom.

Answer 1

The Correct Option is C

Step 1: The energy of a photon emitted during the transition of an electron from \( n = 5 \) to \( n = 1 \) is given by the energy difference between the two states. Using the formula for the energy levels of the hydrogen atom: \[ E_n = -\frac{13.6 \, {eV}}{n^2}. \] The energy of the photon emitted is: \[ \Delta E = E_5 - E_1 = \left(-\frac{13.6}{5^2}\right) - \left(-\frac{13.6}{1^2}\right) = -\frac{13.6}{25} + 13.6 = 13.6 \left( 1 - \frac{1}{25} \right) = 13.6 \times \frac{24}{25} = 13.056 \, {eV}. \] Now converting the energy to joules: \[ E = 13.056 \, {eV} \times 1.602 \times 10^{-19} \, {J/eV} = 2.09 \times 10^{-18} \, {J}. \] Step 2: The recoil energy of the atom will be equal to the energy of the emitted photon. By conservation of momentum, the recoil energy of the hydrogen atom is: \[ E_{{recoil}} = \frac{p^2}{2m_H}, \] where \( p \) is the momentum and \( m_H \) is the mass of the hydrogen atom (\( m_H = 1.67 \times 10^{-27} \, {kg} \)). 
Step 3: The momentum \( p \) is related to the photon energy by \( p = \frac{E}{c} \), where \( c \) is the speed of light. So, \[ E_{{recoil}} = \frac{(E/c)^2}{2m_H} = \frac{(2.09 \times 10^{-18} / 3 \times 10^8)^2}{2 \times 1.67 \times 10^{-27}}. \] Solving this gives: \[ v_{{recoil}} = \sqrt{\frac{2E_{{recoil}}}{m_H}} \approx 4.17 \, {m/s}. \]