Electron Capture by alpha particle -- Frequency of photon?

NucEngMajor
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Homework Statement


Electron with KE = 50eV is captured by Alpha particle, ie. HE++. Calculate the frequency of the emitted photon.

Homework Equations


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

The Attempt at a Solution


Energy before = Energy after
50eV = 4*-13.6eV/1 + hf
f = 1200nm
 
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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.
 
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?
 
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.
 
mfb said:
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.
Would you kindly explain why its T = En + hf and NOT T = U(r) + hf?
 
What is U(r)?
That looks like a very classical approach.
 
mfb said:
What is U(r)?
That looks like a very classical approach.
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?
 
NucEngMajor said:
Initially it is unbound, n = infinity, right?
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.
 
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