X-rays
Practice
practice problem 1
Determine the following quantities for the LCLS…
- the length of a pulse
- the number of wavelengths in a pulse
- the energy of a single x-ray photon in the laser beam
- the energy of a single electron in the linac
- the number of photons produced by one electron assuming 100% efficiency
- the power of a single x-ray pulse if it contains about a trillion photons (n = 1012)
- the average power of the x-ray beam if it pulses 120 times per second
solution
This is a good, simple problem to begin with. Distance is speed times time.
∆s = v∆t
∆s = (3.00 × 108 m/s)(100 × 10−15 s)
∆s = 3.00 × 10−5 mCompare this to the thickness of a piece of paper or the diameter of a human hair. This is a very short distance.
The number of wavelengths in a pulse is the length of a pulse divided by the length of a wave.
n = ∆s λ n = 3.00 × 10−5 m 0.15 × 10−9 m n = 200,000 wavelengths There's one formula for this solution, but because there's two practical units for energy (the joule and the electronvolt) I'll solve it twice.
E = hc λ E = 1.99 × 10−25 J m 0.15 × 10−9 m E = 1.33 × 10−15 J E = hc λ E = 1240 eV nm 0.15 nm E = 8.27 keV These electrons are moving so fast we'll have to use the relativistic formula. Kinetic energy (K) is total energy (E) minus rest energy (E0).
K = E − E0 K = mc2 − mc2 √(1 − v2/c2) K = mc2 ⎛
⎜
⎝1 − 1 ⎞
⎟
⎠√(1 − v2/c2) Again, I'll compute two answers — once for each of the two energy units.
K = (9.11 × 10−31 kg)(3.00 × 108 m/s)2 ⎛
⎜
⎝1 − 1 ⎞
⎟
⎠√(1 − 0.9999999992) K = 1.83 × 10−9 J K = (0.511 MeV) ⎛
⎜
⎝1 − 1 ⎞
⎟
⎠√(1 − 0.9999999992) K = 11.4 GeV Divide the energy of an electron by the energy of a photon to get the maximum number of photons produced by one electron. Use whichever units you wish.
n = Eelectron Ephoton n = (1.83 × 10−9 J) or (11.4 GeV) (1.33 × 10−15 J) (8.27 keV) n = 1,380,000 photons Power in this case is the rate at which energy is transferred. The energy transferred by a pulse is the energy of a single photon times the number of photons in a pulse. We'll have to use the SI units for this part.
Ppulse = Epulse t Ppulse = nphotonsEphoton t Ppulse = (1012)(1.33 × 10−15 J) 100 × 10−15 s Ppulse = 13.3 GW Power is still the rate at which energy is transferred. The energy transferred per second is the energy of a single pulse times the number of pulses in a second. Use the SI units again.
Paverage = Etotal t Paverage = npulsesnphotonsEphoton t Paverage = (120)(1012)(1.33 × 10−15 J) 1 s Paverage = 0.16 W
practice problem 2
solution
Answer it.
practice problem 3
solution
Answer it.
practice problem 4
solution
Answer it.