# Perfectly inelastic collision - energies

1. Apr 2, 2012

### kapitan90

1. The problem statement, all variables and given/known data
An atom of mass M is initially at rest in its ground state. A moving (nonrelativistic) electron of mass $$m_e$$ collides with the atom. The atom+electron system can exist in an 'excited state' in which the electron is absorbed into the atom. The excited state has an extra 'internal' energy E relative to the atom's ground state.

2. Relevant equations
Show that the minimum kinetic energy $$K_{initial}$$ that the electron must have in order to excite the atom is given as:
$$K_{initial} = \frac{(M+m_e)E}{M}$$ and derive a formula for the associated minimum kinetic speed $$v_{0min}$$

3. Additional information
From conservation of momentum $$v_{final} = \frac{m_e v_0}{m_e +M}$$ and so $$KE_{final}=1/2 \frac{(m_e*v_0)^2}{m_e +M}$$
which can be written $$KE_{final}=\frac{K_{initial}}{M+m_e}$$
4. The attempt at a solution

$$minimum KE_{initial} = KE_{final}+ E = \frac{KE_{initial}}{M+m_e} + E$$
so $$KE_{initial}(1-1/(M+m_e))= E$$
$$KE_{initial} = \frac{E}{1-1/(M+m_e)}$$

which is different from the correct answer.
What am I doing wrong?

2. Apr 2, 2012

### Staff: Mentor

Check that last equation.

3. Apr 2, 2012

### kapitan90

I got the mistake, the answer should be:
$$KE_{final}=1/2 (\frac{m_e*v_0}{m_e +M})^2$$
which can be written $$KE_{final}=\frac{K_{initial}m_e}{M+m_e}$$
Hence
$$minimum KE_{initial} = KE_{final}+ E = \frac{K_{initial}m_e}{M+m_e} + E$$
so $$KE_{initial}(1-{m_e}/(M+m_e))= E$$
$$KE_{initial} = \frac{E(M+m_e)}{M}$$

Thank you for your reply!

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