Calculate the velocity of the centre of mass at threshold

[unscientific]

Homework Statement

A high-energy photon γ₁ of energy 261 GeV collides with an infrared photon γ₂ of energy 1eV to produce an electron-positron pair via the reaction

γ₁ + γ₂ → e^- + e⁺

The Attempt at a Solution

Conservation of momentum:

p₁ - p₂ = (0.5)p_e

But since p₁ is 10¹¹ times more than p₂,

p_e ≈ (0.5)(p₁) = 6.96 x 10-¹⁷ kg m s^-1

But when i try to calculate the velocity of the positron/electron using the momentum i get v = c...

Comparing orders of magnitude

p_e ≈ 10^-17

m(γv) = (10^-23) (γ)

This implies that γ ≈ 10⁶...

Am I doing something wrong here?

 

[mfb]

First, working in SI units is a bit impractical here - you get large powers of 10 everywhere.
Second: How can you know that the reaction is at threshold? Otherwise p₁ - p₂ = (0.5)p_e is wrong.

But when i try to calculate the velocity of the positron/electron using the momentum i get v = c...

You should get something really close to, but below c.

This implies that γ ≈ 10⁶...

The order of magnitude is correct.

 

[unscientific]

First, working in SI units is a bit impractical here - you get large powers of 10 everywhere.
Second: How can you know that the reaction is at threshold? Otherwise p₁ - p₂ = (0.5)p_e is wrong.


You should get something really close to, but below c.


The order of magnitude is correct.

I tried to calculate it but my calculator simply gives c...

 

[mfb]

$\frac{v-c}{c} \approx 5 \cdot 10^{-13}$, you would need ~13 digits to see a difference.

 

[unscientific]

$\frac{v-c}{c} \approx 5 \cdot 10^{-13}$, you would need ~13 digits to see a difference.

Is this question a trick question?

 

[mfb]

No, why?
What is the problem statement, by the way? What are you supposed to calculate?

 

[unscientific]

No, why?
What is the problem statement, by the way? What are you supposed to calculate?

It's a 7 marks question, where I'm supposed to find the velocity of the positron/electron pair at threshold energy.

 

[mfb]

Well, you can give the relative deviation from c as result (similar to my post), as the value in m/s is not really interesting.

 

[unscientific]

Well, you can give the relative deviation from c as result (similar to my post), as the value in m/s is not really interesting.

I see, so this is actually an approximation question?

 

[mfb]

I don't understand your question. You can give an exact result, where is the problem?

 

[unscientific]

I don't understand your question. You can give an exact result, where is the problem?

The problem is that my calculator's only able to calculate up to 8 decimal places.. so it can only give 0.99999999c anything beyond that it automatically registers c.

 

[mfb]

You can calculate (v-c)/c with a precision of 8 decimal places then.

 

[unscientific]

You can calculate (v-c)/c with a precision of 8 decimal places then.

Here's what I got; 12 decimal places. When i substituted it into my calculator it gave me a value for gamma as 7*10^5

 

[mfb]

$\beta = \sqrt{1-10^{-12}} \approx 1-\frac{1}{2}10^{-12}$
$1-\beta \approx 5 \cdot 10^{-13}$
This allows to determine (1-β) with the same relative precision as γ.