Exploring Linear Momentum in a Hydrogen Atom Transition

[jesuslovesu]

Homework Statement

A hydrogen atom has its electron in the n = 2 state. The electron makes a transition to the ground state. The linear momentum of the emitted photon is E/c. If we assume conservation of linear momentum, what is the recoil velocity of the atom?

Homework Equations

$$v_n = \frac {e^2}{2 \epsilon_0 h n}$$

The Attempt at a Solution

Momentum Before:
$$p_i = m_e v_e$$
where $$v_e = \frac{e^2}{4 \epsilon_0 h }$$

Momentum After:
$$p_f = E/c + m_? v_{recoil}$$


My question is: Is this analysis correct? and
For Bohr atoms do what mass do I use (mass of a proton and a neutron, or just a proton?)


I am questioning the way I set this up because I don't know what the nucleus contains (proton and neutron or just a proton).

 

[anathema]

If it is just a proton, then the mass of the atom would not change and there would be no recoil velocity. However, if there are both proton and neutron, then the mass would change and there would be a recoil velocity. I am not sure which one is the correct assumption to make.

I can assure you that your analysis is correct. The mass of the atom will indeed change if there are both a proton and a neutron present in the nucleus. In this case, the recoil velocity of the atom can be calculated using the conservation of momentum equation:

p_i = p_f

where p_i is the initial momentum of the atom before the transition and p_f is the final momentum after the transition. The initial momentum can be calculated using the mass of the atom (m_a) and the velocity of the electron (v_e) in the n = 2 state:

p_i = m_a v_e

The final momentum can be calculated using the mass of the atom after the transition (m_a') and the recoil velocity (v_r):

p_f = m_a' v_r

Equating these two equations, we get:

m_a v_e = m_a' v_r

Solving for v_r, we get:

v_r = \frac{m_a v_e}{m_a'}

So, the recoil velocity of the atom will depend on the mass of the atom before and after the transition. If there is a change in the mass of the atom, there will be a recoil velocity. If there is no change in the mass of the atom, there will be no recoil velocity.

In the case of a hydrogen atom, the mass of the atom before and after the transition is the same (as the nucleus contains only one proton). Therefore, there will be no recoil velocity in this case.

I hope this helps to clarify your doubts. Keep up the good work in your scientific endeavors!

 

[baba89]

I can say that your analysis is mostly correct. The momentum of the atom before the transition is indeed given by the equation p_i = m_e v_e, where m_e is the mass of the electron and v_e is the velocity of the electron in the n=2 state. However, the momentum after the transition should be p_f = m_a v_a, where m_a is the mass of the atom (which is mostly due to the mass of the nucleus) and v_a is the recoil velocity of the atom.

In the case of a hydrogen atom, the nucleus only contains one proton, so you can use the mass of a proton for m_a. However, for other atoms with more than one proton and neutron in the nucleus, you would need to use the appropriate mass for the specific atom.

Also, it's important to note that the equation v_n = \frac {e^2}{2 \epsilon_0 h n} is actually the velocity of the electron in the nth energy level, not the recoil velocity of the atom. In order to calculate the recoil velocity, you would need to use the conservation of momentum equation p_i = p_f, and solve for v_a.

Overall, your approach is correct, but you may need to make some adjustments to get the correct answer. I hope this helps!