# Pair Production Physics Question

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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

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

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#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.

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#3

(Original post by

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

**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

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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?

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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

this is what I got

**Yazzy.j**)this is what I got

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#7

(Original post by

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?

**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?

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#9

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.

2

*E*_{γ}= 2*mc*^{2}+ ½*mv*^{2}
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#10

(Original post by

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.

**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.

2

*E*_{γ}= 2*mc*^{2}+ ½*mv*^{2}
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#11

(Original post by

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?

**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?

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#12

**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

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Your tutor is wrong- i managed to figure out my mistake- see attached for correct workings

(Original post by

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

**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

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#14

(Original post by

Your tutor is wrong- i managed to figure out my mistake- see attached for correct workings

**ben.alexander**)Your tutor is wrong- i managed to figure out my mistake- see attached for correct workings

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