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atom emits photon

a hydrogen atom at rest emits a photon of energy 20 eV, what is the resulting velocity of the atom?

don't you say that Ek(->)=Ek(<-). then equate the energy of the photon with the resulting Ek of the atom, and things like that?

Reply 2

john!!

the atom and photon dont nessecarily have the same kinetic energy

Reply 3

Robob

can't you just do it by using the conservation of linear momentum?

momentum of photon = E/c

therefore momentum of atom = -E/c

and you can do it from there

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Oh, is this ignoring relativistic effects?

Reply 4

john!!

E=pc doesn't apply to massive bodies, for the atom, p = gamma*mv

Reply 5

Robob

john!!

E=pc doesn't apply to massive bodies, for the atom, p = gamma*mv

You mistunderstood my post, the momentum of the atom will have the same magnitude as that of the emitted photon, but in the opposite direction, hence momentum of the atom = -E/c.

You then equate p = gamma*mv and p = -E/c to find v

But seeing as it was your work I didn't want to do the entire thing for you

Reply 6

john!!

would you mind giving me an answer that I can check against my own?
thanks for the help but I do know about momentum conservation

Reply 7

Ana

john!!

would you mind giving me an answer that I can check against my own?
thanks for the help but I do know about momentum conservation

How did you find the exam?

Reply 8

john!!

hard! some questions were fine but there was one, which gave the width of some random particle in MeV/c^2, what?? that isnt even a length. Thats like saying "I've been waiting at this bus stop for 5 volts". Apart from that one, good. how about you?

Reply 9

Ana

john!!

Not too bad. Couldn't do the lifetime of the meson, nor explanation of K_alpha X-rays (guess who did their revision at the last minute :redface:

), but feel pretty good about the rest. I did Q2 and Q3. Q3 was a piece of fish, it just boiled down to some messy simultaneous equations.

Reply 10

john!!

yeah. Q3 was basically A level physics I thought, nice and simple.

the funny thing is, in question two for the second 7 mark question I forgot that delta t' was zero so I made a quadratic in v and used the formula to solve it. It was so messy (and gave me half the answer I wanted because I'd written down half the time period), which wasted about 15 minutes. Then I went back and did it in like 2 lines. Stupid.

But now its all over, now Im free so it doesnt matter!!!

Reply 11

Ana

john!!

Out of interest, how did you do that bit?

Reply 12

john!!

x = 1000
t = 0.000002
t' = 0

t' = gamma(t - vx/cc)
0 = gamma(t - vx/cc)
t = vx/cc
v = tcc/x = 0.6c I got. Its much simpler than I first thought.

Reply 13

Ana

john!!

x = 1000
t = 0.000002
t' = 0

t' = gamma(t - vx/cc)
0 = gamma(t - vx/cc)
t = vx/cc
v = tcc/x = 0.6c I got. Its much simpler than I first thought.

Bother. I got 0.3c, by using the Lorentz transformations.

Reply 14

AlphaNumeric

john!!

Many relativistic or quantum mechanical quantities are given in what appear to be incorrect units. This is because it's usual to set things like

\hbar

, c and G equal to 1. Doing this means you end up turning all your units into powers of Joules.

Certainly has it's uses when you don't have to keep track of powers of c and

\hbar

all the way through what are nasty enough quantum mechanical integrals or relativistic expansions.

Once you have your answer, you can then reconvert back to correct units without much bother.