How much energy is required to boost an electron to 0.9c, 0.99c, 0.999c? What is E?
(6) A particle moves at a speed such that its kinetic energy just equals its rest mass energy.
How fast is the particle moving?
(7) An electron in a certain x-ray tube is accelerated from rest through a potential difference of
180,000 volts in going from the cathode to the anode. When it arrives at the anode what is its
(a) kinetic energy in eV? (b) its "relativistic mass" mg? (c) its relativistic velocity? (d) (the value
of e/m??) (e) its velocity calculated classically?
(8) Calculate the amount of work in MeV that must be done (a) to bring an electron from rest to
a velocity of 0.4 c; and (b) to increase its velocity from 0.4 c to 0.8 c . (c) What is the ratio of the
kinetic energy of the electron at the velocity of 0.8 c to that at 0.4 c when computed from (1)
relativistic values and (2) from classical values?
(9) Each stage of a three-stage rocket can boost the stage (plus cargo) to 0.9c. What is the
ultimate velocity of the rocket and how long would it take to get to alpha centauri (4.3 light years
distant)?
(10) What can you say about the reaction γ → e+e-? ? (Show me!)
(11) Sure, the γ (problem 10 above) needs a nucleus to absorb the momentum. Can you show the nucleus must recoil by
p(nucleus)=2me²c²/pg ???
OK, some hints here....first, note this problem is for all energies, not just the threshold energy.
Next, note the recoil actually decreases with increasing photon energy.
Hmmm.... Seems that the electron/positron pair could almost carry away the momentum by themselves, and the higher the energy the closer they come to it....oh yes, look at the formula, no nucleus mass? must be large (proton is 1836 electron masses, nuclei are even more massive) and that is the key: The nucleus just has to carry off a teensy bit of the momentum and so hardly takes any of the photon E away. Try using energy conservation with this in mind, let the photon change into the electron+positron p=2mgc, get the speed, and the nucleus must absorb the difference between the photon p and the 2mgv...
Finally, p(nucleus)<<p(e) -- the nucleus hardly does anything more than provide a frame of reference! (Well... actually the large mass is responsible for this but there is actually another fundamental reason for the nucleus being responsible for the pair production to occur, and that is the nuclear charge. The probability of pair production occurring at all is proportional to Z².)
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Hmmmm.... if you are just at the threshold energy you find p(N)pg=4m²c² ... it goes from 4 to 2 as the electrons go from v<<c (threshold) to v~c (high energy limit). You will get a term c³/(c-v) and that's why. (Electrons don't require much energy to make the 2 valid. Answer.)
(12) An Omega particle (mass=1672 MeV/c²) at rest decays into a lambda particle (1115.6 MeV/c²) and a Kaon (493.7 MeV/c²). What are the energies of the lambda and Kaon?
....Hint...Try playing with E²=(pc)²+(mc²)² ......this is easy!
Algebra lesson:
If A = SQRT(X²+B²) + SQRT(X²+C²)
then it is easy to show that
X² = 0.25*(A²+B²-C²)²/A² - B²
or
X² = 0.25*(A²-B²+C²)²/A² - C²
That is useful for "A decays to B & C" ---A,B,C are the mc², X² is the (pc)²....
Just move one of the square roots to the LHS and square both sides (if you don't do this you get a mess).
A² + (X²+B²) -2A SQRT(X²+B²) = X²+C²
Ha! Cancel the X², rearrange:
SQRT(X²+B²) = (A² + B² - C²)/2A and square again... there you are!