Photon Absorption by Hydrogen Molecule: Velocity and Translational Energy Calculation

Question:

When a hydrogen molecule absorbs a photon from a helium gas discharge tube, what is the velocity of the hydrogen molecule after absorbing the photon? Also, what is the translational energy of the hydrogen molecule in Jmol-1?

Answer:

The velocity of the hydrogen molecule after absorbing the photon is approximately 342 m/s, and the translational energy of the hydrogen molecule is approximately 1.68 x 10³ Jmol-1.

Final Answer:

When a hydrogen molecule absorbs a photon, it gains velocity due to the photon's momentum. The velocity of the hydrogen molecule is approximately 342 m/s. The translational energy of the hydrogen molecule is approximately 1.68 x 10³ Jmol-1.

Explanation:

To solve this, we need to use the principles of conservation of momentum, energy and the properties of photons.

First, the momentum of the absorbed photon is calculated by p = h/λ, where h is Planck's constant and λ is the wavelength. With λ = 58.4nm and h = 6.626 x 10⁻³⁴ Js, we get p ≈ 1.1345 x 10⁻²⁷ kg m/s.

Assuming no initial momentum, the final momentum of the hydrogen molecule must equal this value, so we can calculate the velocity: v = p/m, where m is the mass of the hydrogen molecule. The molecular mass of hydrogen is about 3.32 x 10⁻²⁷ kg which gives v ≈ 342 m/s.

Lastly, we use the kinetic energy formula (K.E. = 1/2 mv²) with the given velocity to find the translational energy per molecule. To convert to Jmol-1, we multiply by Avogadro's number (NA = 6.022 x 10²³ mol⁻¹) which gives an energy of about 1.68 x 10³ Jmol-1.