# Relativity question

1. Apr 13, 2005

### quasar987

A photon of energy $h\nu$ collides with a still atom in an excited state. After the collision, the photon has the same energy $h\nu$, but its direction is now antiparallel to its initial direction. The atom is now in its fundamental state. What was the energy of the state of the atom?

What I did thus far:

The conservation of momentum statement is

$$h\nu/c = -h\nu/c + p_{af} \Rightarrow p_{af} = 2h\nu/c$$

The conservation of energy statement is

$$h\nu + m_a_ic^2 = h\nu + m_a_fc^2 + K_a_f \Leftrightarrow m_a_ic^2 = m_a_fc^2 + K_a_f$$

(a is for atom, i is for initial, f is for final, K is the kinetic energy)

For these to be equal, the mass of the atom before and after collision has to be different. This change of mass has to be associated with a change of energy of the atom's constituants. So I figured the energy associated with this change of mass is the energy corresponding to the state switch. In other words, the answer to the question is that the energy of the initial state of the atom is given by

$$(m_a_i - m_a_f)c^2$$

But I don't know how to find this. From

$$E_{af}^2 = p_{af}^2c^2+m_{af}^2c^4 = m_{ai}^2c^4 = E_{ai}^2$$

I can only get

$$m_{ai}^2-m_{af}^2 = 4h^2\nu^2/c^4$$