To find the atom's velocity after absorbing the photon, we can make use of the conservation of energy and momentum. Since the collision is inelastic, we assume that the atom absorbs the entire energy of the photon.
Let's start by calculating the momentum of the photon. The momentum of a photon can be found using the equation:
p = E/c
Where p is the momentum, E is the energy of the photon, and c is the speed of light (approximately 3.00 × 10^8 m/s).
Plugging in the values: E = 10.2 eV = 10.2 × 1.6 × 10^-19 J (1 eV = 1.6 × 10^-19 J) c = 3.00 × 10^8 m/s
p = (10.2 × 1.6 × 10^-19 J) / (3.00 × 10^8 m/s) p ≈ 5.44 × 10^-28 kg·m/s
According to the conservation of momentum, the total momentum before the collision should be equal to the total momentum after the collision. Initially, the hydrogen atom is assumed to be at rest, so its momentum is zero.
Since the collision is inelastic, the atom will gain momentum after absorbing the photon. Let's denote the velocity of the atom after the collision as v. The momentum of the atom can be calculated using:
p_atom = m_atom × v
Where p_atom is the momentum of the atom, m_atom is the mass of the atom, and v is the velocity of the atom.
Equating the momentum before and after the collision:
0 + p = m_atom × v
Solving for v:
v = p / m_atom v = (5.44 × 10^-28 kg·m/s) / (1.67 × 10^-27 kg) v ≈ 0.325 m/s
Therefore, the velocity of the hydrogen atom after absorbing the photon is approximately 0.325 m/s.