# Pair Production Physics QuestionWatch

Announcements
#1
Can someone produce a quick worked solution of this question? I keep getting a negative value of wavelength.
A stationary neutron is annihilated by an anti-neutron that has a velocity of 9.0 x 107 ms-1. Calculate the wavelength of the gamma photons produced.

Here's my working:

Erest(neutron)= 939.551E6 x 1.6E-19= 1.50E-10 J

Ek(antineutron)= 1/2mv^2 => Ek= ½ 1.675E-27 (9E7)^2 =6.784E-9 J

E(photon)=hc/lamda => lamda=hc/E= 6.63E-34x3E8/(1.5E-10-6.784E-9)= -2.998E-17 m
0
6 months ago
#2
From what I can remember, the total energy in the system is the rest energies of the two particles plus the kinetic energy of the anti-neutron. Add all three energies together to get the total. On collision, this energy is then released as two photons each with half the total energy. From this you can then work out the energy of each photon and hence, the wavelength.
0
6 months ago
#3
(Original post by ben.alexander)
Can someone produce a quick worked solution of this question? I keep getting a negative value of wavelength.
A stationary neutron is annihilated by an anti-neutron that has a velocity of 9.0 x 107 ms-1. Calculate the wavelength of the gamma photons produced.

Here's my working:

Erest(neutron)= 939.551E6 x 1.6E-19= 1.50E-10 J

Ek(antineutron)= 1/2mv^2 => Ek= ½ 1.675E-27 (9E7)^2 =6.784E-9 J

E(photon)=hc/lamda => lamda=hc/E= 6.63E-34x3E8/(1.5E-10-6.784E-9)= -2.998E-17 m this is what I got
0
6 months ago
#4
0
#5
I thought that the total energy would be the energy of the neutron. Therefore the energy of the photon emitted is the rest energy of neutron - kinetic energy of the anti neutron. Why is this incorrect?
0
#6
But that only takes into account the energy of the anti-neutron, not the energy of the neutron it collided with- how can this be right?

(Original post by Yazzy.j) this is what I got
0
6 months ago
#7
(Original post by ben.alexander)
But that only takes into account the energy of the anti-neutron, not the energy of the neutron it collided with- how can this be right?

Yeah I just realised that mistake. I messaged my tutor with that question and hopefully he'll answer soon. If he doesn't reply by tonight, maybe just ask your teacher tomorrow.
1
6 months ago
#8 my tutor worked it out. here you go
1
6 months ago
#9
(Original post by Yazzy.j) my tutor worked it out. here you go

I disagree that the energy of the photon comes from the kinetic energy of the moving anti-neutron.

The energy of the 2 photons comes from the sum of the kinetic energy of the anti-neutron and rest-mass energy of the neutron and anti-neutron.
2Eγ = 2mc2 + ½mv2
0
6 months ago
#10
(Original post by Eimmanuel)
I disagree that the energy of the photon comes from the kinetic energy of the moving anti-neutron.

The energy of the 2 photons comes from the sum of the kinetic energy of the anti-neutron and rest-mass energy of the neutron and anti-neutron.
2Eγ = 2mc2 + ½mv2
yeah that's what I was thinking. I'm gonna ask my teacher tomorrow coz I don't think I agree with what my tutor said. I'll post on this chat what he says tomorrow morning if I remember
0
6 months ago
#11
(Original post by ben.alexander)
I thought that the total energy would be the energy of the neutron. Therefore the energy of the photon emitted is the rest energy of neutron - kinetic energy of the anti neutron. Why is this incorrect?
The total energy is the sum of the kinetic energy of the anti-neutron and rest-mass energy of the neutron and anti-neutron. Both neutron and anti-neutron are annihilated after the collision to produce a pair of identical photons.
0
6 months ago
#12
(Original post by ben.alexander)
Can someone produce a quick worked solution of this question? I keep getting a negative value of wavelength.
A stationary neutron is annihilated by an anti-neutron that has a velocity of 9.0 x 107 ms-1. Calculate the wavelength of the gamma photons produced.

Here's my working:

Erest(neutron)= 939.551E6 x 1.6E-19= 1.50E-10 J

Ek(antineutron)= 1/2mv^2 => Ek= ½ 1.675E-27 (9E7)^2 =6.784E-9 J

E(photon)=hc/lamda => lamda=hc/E= 6.63E-34x3E8/(1.5E-10-6.784E-9)= -2.998E-17 m
Teenie2 had outlined the correct way of attempting the problem in post #2.
0
#13
Your tutor is wrong- i managed to figure out my mistake- see attached for correct workings

(Original post by Yazzy.j)
yeah that's what I was thinking. I'm gonna ask my teacher tomorrow coz I don't think I agree with what my tutor said. I'll post on this chat what he says tomorrow morning if I remember
0
6 months ago
#14
(Original post by ben.alexander)
Your tutor is wrong- i managed to figure out my mistake- see attached for correct workings
Agreed. 0
X

new posts Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### University open days

• University of Hull
Sat, 23 Nov '19
• Edge Hill University
Sat, 23 Nov '19
• Keele University
Sat, 23 Nov '19

### Poll

Join the discussion

#### What offers have you received from universities?

Unconditional (53)
21.03%
Unconditional if firmed (20)
7.94%
Unconditional if insurance (2)
0.79%
Both unconditional and unconditional if firm/insurance (6)
2.38%
No unconditional offers (171)
67.86%