X‑rays

Practice

practice problem 1

The US Department of Energy's Linac Coherent Light Source (LCLS) in Menlo Park, California was the world's most powerful x-ray laser when it opened for business in 2009. The LCLS generates x-rays from a beam of electrons accelerated from zero to 99.9999999 per cent the speed of light in 3.2 km. The central tube of this linear accelerator or linac is the straightest object in the world and lies beneath the longest building in the United States. The beamline ends at a 100 m long series of alternating magnets called an undulator where the electrons slalom back and forth in a synchronized fashion generating coherent x-rays. The pulses are quick enough (∆t < 100 fs) that ultrafast events can be captured and the x-rays are short enough (λ = 0.15 nm) that ultrasmall objects can be imaged. By stringing together sequences of ultrafast, ultrasmall images the LCLS is effectively a movie camera for individual molecules.

Determine the following quantities for the LCLS …

  1. the length of a pulse
  2. the number of wavelengths in a pulse
  3. the energy of a single x-ray photon in the laser beam
  4. the energy of a single electron in the linac
  5. the number of photons produced by one electron assuming 100% efficiency
  6. the power of a single x-ray pulse if it contains about a trillion photons (n = 1012)
  7. the average power of the x-ray beam if it pulses 120 times per second

solution

  1. This is a good, simple problem to begin with. Distance is speed times time.

    s = vt = (3.00 × 108 m/s)(100 × 10−15 s) = 3.00 × 10−5 m

    Compare this to the thickness of a piece of paper or the diameter of a human hair. This is a very short distance.
  2. The number of wavelengths in a pulse is the length of a pulse divided by the length of a wave.
             
    n =  s  =  3.00 × 10−5 m  = 200,000
    λ 0.15 × 10−9 m
             
  3. There's one formula for this solution, but because there's two practical units for energy (the joule and the electron volt) I'll solve it twice.
             
    E =  hc  =  1.99 × 10−25 J m  = 1.33 × 10−15 J
    λ 0.15 × 10−9 m
             
    E =  hc  =  1240 eV nm  = 8270 eV
    λ 0.15 nm
             
  4. 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 =  mc2  − mc2 = mc2  ⎛
    ⎝    		 1  − 1  ⎞
    ⎠
    √(1 − v2/c2) √(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  ⎞
    ⎠    		  = xxxxxxx J
    √(1 − 99.99999992)
     
    K = (0.511 × 106 eV)  ⎛
    ⎝    		 1  − 1  ⎞
    ⎠    		  = xxxxxxx eV
    √(1 − 99.99999992)
     
  5. 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.
                 
    nphoton =  Eelectron  =  (xxxxxxx)  =  (xxxxxxx)  = xxx
    Ephoton (xxxxxxxx) (xxxxxxxx)
                 
  6. 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  =  nphotonsEphoton  =  (1012)(xxxxxxxx)  = xxxxxxxx
    t t 100 × 10−15 s
                 
  7. 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  =  npulsesEpulse  =  (120)(1012)(xxxxxxxx)  = xxxxxxxx
    t t 1 s
                 

practice problem 2

Write something.

solution

Answer it.

practice problem 3

Write something.

solution

Answer it.

practice problem 4

Write something completely different.

solution

Answer it.

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