Calculate energy and momentum of a hydrogen atom in a photon emission transition

[rayman123]

Homework Statement

'
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

Homework Equations

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

The Attempt at a Solution

having calculated the energy of the emitted photon i can calculate its momentum
$$p= \frac{E_{photon}}{c}=\frac{2.04\cdot10^{-18}}{3\cdot10^8}=6.81\cdot10^{-27}\frac{kg\cdot m}{s}$$

Homework Statement

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

 

[Redbelly98]

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.

 

[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?

 

[Redbelly98]

Looks good!

Sorry for not replying sooner, somehow I missed the notification that you had posted again.

 

[rayman123]

thank you!;)