Problem Solving Strategies

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John Rennie

5:02 AM

@mechanist Ah, OK, yes I agree. I'd missed the fact it was a dot product.

2 hours later…

mechanist

7:30 AM

But then differentiating it should make the same statement true for acceleration true, but it isn't.

Harjot Dhillon

8:25 AM

According to the solution entire KE of photon get transferred to photon

But why can't we solve it by assuming entire momentum to be transferred to photon

After collision of electron with heavy nucleus they would be nearly on rest and hence we can assume their momentum to be zero.

electron in first line instead of photon in KE of photon get

Harjot Dhillon

9:13 AM

@JohnRennie hello sir

John Rennie

Hi :-)

Harjot Dhillon

Could you have a look at this question

John Rennie

I don't think you can assume anything about the momentum because the collision of the electron with the target is a complicated process involving many bodies.

Harjot Dhillon

I know, but assuming anything about KE would also be complicated. So how can we prove that entire KE get transferred

John Rennie

KE is a scalar so we can write:
E(electron) = E(photon) + E(target)
where all the energies are scalars and we just add the magnitudes.

So we get the maximum value of E(photon) when E(target) = 0

And in that case we just have;

E(electron) = E(photon)
which is what the solution uses.

OK so far?

Harjot Dhillon

9:20 AM

Yes

John Rennie

Now let's consider momentum. As before we have:
p(electron) = p(photon) + p(target)
but now these are vectors. If we take the direction of the electron as +ve then we expect p(target) to be positive but p(photon) will be -ve because the photon is heading away from the target in the opposite direction to the electron.
Yes?

Harjot Dhillon

Yes

John Rennie

So we should write the equation as:
p(electron) = p(target) - |p(photon)|

But suppose we set p(target) = 0, then we'd have a positive value on the left side and a negative value on the right side and the equality is impossible.

We must have |p(target)| > |p(photon)|
Yes?

Harjot Dhillon

So photon always go in opposite direction

I didn't knew this before

John Rennie

Well it's more complicated than that as I think the photons typically come out of the x-ray tube at an angle not exactly opposite to the electrons. But the point is that because we are now adding vectors it's a lot more complicated than just adding energies.

Harjot Dhillon

9:30 AM

But why can't photon move in same direction as electron.

John Rennie

Because the photon is always emitted in the opposite direction to the acceleration vector.

Harjot Dhillon

Now I see where we are going

I understood now

We must have |p(target)| > |p(photon)| and by applying conservation of energy we can prove that photon momentum can't exceed that of electron

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