# Homework Help: Electron Capture by alpha particle -- Frequency of photon?

1. Nov 10, 2015

### NucEngMajor

1. The problem statement, all variables and given/known data
Electron with KE = 50eV is captured by Alpha particle, ie. HE++. Calculate the frequency of the emitted photon.

2. Relevant equations
KE = m/2 v^2; E=hf, En = Z^2*-13.6eV/n^2

3. The attempt at a solution
Energy before = Energy after
50eV = 4*-13.6eV/1 + hf
f = 1200nm

2. Nov 10, 2015

### Staff: Mentor

Something went wrong between the last two lines, the wavelength is way too long.
Also, 1200 nm is a length, not a frequency.

The electron doesn't have to go to the ground state directly, so you should at least say that you assume this.

3. Nov 10, 2015

### NucEngMajor

Right. I solved for wavelength. 50 eV is a big KE so 1200nm seems reasonable? In general is the conservation of energy correct? I mean is it appropriate to put En + hf rather than Potential + hf?

4. Nov 10, 2015

### Staff: Mentor

The approach is right, just the result is wrong.
1200 nm is infrared light, with an energy of about 1 eV per photon. And 1200 nm is not a frequency, the problem statement asks for the frequency.

5. Nov 10, 2015

### NucEngMajor

Would you kindly explain why its T = En + hf and NOT T = U(r) + hf?

6. Nov 10, 2015

### Staff: Mentor

What is U(r)?
That looks like a very classical approach.

7. Nov 11, 2015

### NucEngMajor

I'm just confused why the energy is En plus some additional hf. Initially it is unbound, n = infinity, right? After we drop to some E1 or En?

8. Nov 11, 2015

### Staff: Mentor

It is unbound with its energy above zero, so "n=infinity" is a problematic concept.

Initially it is unbound and has a positive energy, afterwards it is bound and has a negative energy. The energy of the photon is the difference of the two states. Subtracting a negative value from a positive will lead to an answer that is larger than the positive value. In other words, your photon will have an energy of more than 50 eV.