# Recoil energy and velocity

1. Sep 4, 2010

### rayman123

1. The problem statement, all variables and given/known data'
Calculate the recoil energy and velocity of a hydrogen atom in a transition from n=4 to n=1 in which a photon is emitted

2. Relevant equations

I started with calculating the energy of the emitten photon.
$$E_{photon}= \frac{-16.6 eV}{n^2}-\frac{-13.6 eV}{n^2}=\frac{-13.6}{16}+13.6= 12.75 eV$$

3. The attempt at a solution
having calculated the energy of the emitted photon i can calculate its momentum
$$p= \frac{E_{photon}}{c}=\frac{2.04\cdot10^{-18}}{3\cdot10^8}=6.81\cdot10^{-27}\frac{kg\cdot m}{s}$$
1. The problem statement, all variables and given/known data

can this be relevant? can the emitted photons momentum be so small??? how to solve this problem?

2. Sep 4, 2010

### Redbelly98

Staff Emeritus
Yes, the photon's momentum is really that small.

Has your class discussed any conservation laws lately? You can think of this as a collision-like problem involving the hydrogen atom and the photon.

3. Sep 4, 2010

### rayman123

ah okej,i might have mixed up the velocity with the momentum but know i know
Yes we had classes in conservation laws and i remember from it that in an isolated system with only two objects, the change in momentum of one object must be equal and opposite to the change in momentum of the other object so that could mean that momentum of the photon is the same as the momentum of the hydrogen atom is that right?

in that case non-relativistic momentum for the hydrogen atom will be
$$p= m\cdot v$$ where m-the hydrogen atom mass, v-its velocity

from this i can calculate the velocity of the hydrogen atom
$$v=\frac{p}{m}= \frac{6.81\cdot 10^{-27}}{1.67\cdot 10^{-27}} = 4.078\frac{m}{s}$$

recoil energy of the hydrogen atom will be
$$E_{recoil}= \frac{mv^2}{2}=\frac{1.67\cdot10^{-27}\cdot(4.078)^2}{2}=1.39\cdot10^{-26} J$$

is that a correct solution?

Last edited: Sep 4, 2010
4. Sep 6, 2010

### Redbelly98

Staff Emeritus
Looks good!