Is it Possible for a Photon to Create an Electron-Positron Pair?

[thata6]

an electron-positron pair, each with a kinetic energy of 220 Kev, is produced by a photon. find the
energy and wavelength of the photon.

started with E=hf=pc , and the fact that p=mv, but his not working

 

[thata6]

never mind I solved this using
total E = rest E + KE
E = Mc2 + 220Kev double this since we have e- and e+ and you have the protons E
then use E=hf to find wavelength.

 

[turin]

an electron-positron pair, each with a kinetic energy of 220 Kev, is produced by a photon. find the
energy and wavelength of the photon.

... use E=hf to find wavelength.

Be careful. How do you find wavelength from this? It may seem like a silly question, but actually, you can't even solve this problem without more information or assumptions (i.e. the relative directions of the electron and positron) that have no reason to be true in general. BTW, did you mention a proton, or was that a typo?

 

[thata6]

hi there.
ok let me step you through my thinking
1.the energy of the photon will be the energy of the positron plus the energy of the electron

energy of e- = rest energy + kinetic energy
= Mc2 + 3.52x10^-14 (220 Kev)
= 8.1791x10^-14J + 3.52x10^-14J

 

[thata6]

hi there.
ok let me step you through my thinking
1.the energy of the photon will be the energy of the positron plus the energy of the electron

energy of e- = rest energy + kinetic energy
= Mc2 + 3.52x10^-14 (220 Kev)
= 8.1791x10^-14J + 3.52x10^-14J
X2 since we have e- and p+
= 2.33982x10^-13J = 1.46239 Ev
is the total energy that the proton must of had in order to create this pair


the E = hc x wavelength^-1 (dont know how to do fractions here)
substitute values and
wavelength is is 8.494x10^-13 M

what do you think?

 

[turin]

1.the energy of the photon will be the energy of the positron plus the energy of the electron

energy of e- = rest energy + kinetic energy

Yes. Momentum must also be conserved.

the E = hc x wavelength^-1 (dont know how to do fractions here)
...
what do you think?

What do you think c is in this case. The actual equations that you used here were
<br /> E=h\nu<br />
and
<br /> u=\nu\lambda<br />
where you have assumed that u=c. However, the correct equation to use for the wavelength is
<br /> p=h\lambda<br />
Unfortunately, your problem statement does not give enough information to determine the momentum of the photon, so the best you can do is to come up with an upper limit on the momentum (and thus a lower limit on the wavelength).

For example, if the electron and positron are produced so that they travel in opposite directions, what must be the momentum of the photon? And, therefore, what must be its wavelength?

 

[Avodyne]

The problem here is that this process is impossible; a real photon cannot convert itself to an electron-positron pair. If the problem refers to a virtual photon, then turin is correct that there is not enough information.

 

[turin]

The problem here is that this process is impossible; a real photon cannot convert itself to an electron-positron pair. If the problem refers to a virtual photon, then turin is correct that there is not enough information.

This process is not impossible. The problem is that the OP is using the wrong physical principle. The OP did not specify that this photon was "real", and, as you point out, it cannot even be real anyway. But at any rate I personally hate that terminology. Every photon is just as real as every other photon, regardless of how far from its mass-shell it is.