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Physics 7B Homework Set 10 Solutions: Momentum-Energy Diagrams and Atomic Energy Levels, Study notes of Physics

Physics

Solutions to homework set 10 of physics 7b, including calculations and diagrams for momentum-energy diagrams of particles and atomic energy levels. Topics covered include particle motion, rest energy, kinetic energy, total energy, relativistic momentum, velocity, and wavelengths of photons in atomic transitions.

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Physics 7B Spring 2003
Homework Set 10 solutions
1. Page 886, Problem 17. Add: Illustrate the motion of such a particle by sketching a
momentum-energy diagram (a plot of Evs pc) with an appropriate vector to represent this
particle. You may assume the mass of such a particle to be m, and so be sure to include in
your sketch the “hyperbola of mass-mparticles”.
p=mv
p1−v2/c2= 3mv so 1 −v2/c2= 1/9,or
v
c=√8
3=2√2
3= 0.9428
Below is a momentum-energy diagram that illustrates such a particle. [Note that
E−mc2=K= 2mc2, illustrated correctly on this diagram.] The hyperbola of mass-m
particles is shown: E2−p2c2=m2c4, and the vector representing our particle has its
arrowhead ending on this hyperbola.
mc2
3mc2
v
c
_
θtan ==E
pc
_= 0.9428
pc
E
0
θ
2. Consider an electron. (a) What is the rest energy of this electron, both in Joules and
keV? (b) If this electron were accelerated by a potential difference ∆V= 100 kilovolts
(105volts), what would its kinetic energy Kbe, both in Joules and keV? (c) What would
its total energy Ebe, both in Joules and keV? (d) What would its (relativistic) momentum
be, both in kilogram-meters per second, and in keV/c? [Ans: 335 keV/c] (e) What
would its velocity be, both in meters per second, and as a fraction of c? (f) Sketch a
momentum-energy diagram showing the motion of this electron, with both the energy axis
pf3

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Physics 7B Spring 2003

Homework Set 10 solutions

  1. Page 886, Problem 17. Add: Illustrate the motion of such a particle by sketching a momentum-energy diagram (a plot of E vs pc) with an appropriate vector to represent this particle. You may assume the mass of such a particle to be m, and so be sure to include in your sketch the “hyperbola of mass-m particles”.

p =

mv √ 1 − v^2 /c^2

= 3mv so 1 − v^2 /c^2 = 1/ 9 , or

v c

Below is a momentum-energy diagram that illustrates such a particle. [Note that E − mc^2 = K = 2mc^2 , illustrated correctly on this diagram.] The hyperbola of mass-m particles is shown: E^2 − p^2 c^2 = m^2 c^4 , and the vector representing our particle has its arrowhead ending on this hyperbola.

mc^2

3mc 2

v c tan θ = __ = E_

pc_ = 0.

pc

E

θ

  1. Consider an electron. (a) What is the rest energy of this electron, both in Joules and keV? (b) If this electron were accelerated by a potential difference ∆V = 100 kilovolts (10^5 volts), what would its kinetic energy K be, both in Joules and keV? (c) What would its total energy E be, both in Joules and keV? (d) What would its (relativistic) momentum be, both in kilogram-meters per second, and in keV/c? [Ans: 335 keV/c] (e) What would its velocity be, both in meters per second, and as a fraction of c? (f) Sketch a momentum-energy diagram showing the motion of this electron, with both the energy axis

E and the momentum axis pc scaled in keV. The moving electron should be represented by a vector on this diagram, and it should include (as in the previous problem) the appropriate hyperbola. It should also illustrate the magnitudes of E, K and pc.

(a) The rest energy mc^2 = (9. 109 × 10 −^31 )(8. 99 × 1016 ) = 8. 189 × 10 −^14 Joules, or

  1. 11 × 105 eV, or 511 keV.

(b) If the electron is accelerated by 100 kV, it will have a kinetic energy K = 100 keV = 1. 6 × 10 −^14 Joules.

(c) The total energy of the electron is E = mc^2 + K = 611 keV, or 9. 78 × 10 −^14 Joules.

(d) The momentum of the electron is most easily found from the equation for the invariant hyperbola: E^2 − p^2 c^2 = m^2 c^4 , or

p^2 c^2 = E^2 − m^2 c^4 = 611^2 − 5112 = 112200 (keV)^2 so pc =

112200 = 335 keV

and therefore p = 335 keV/c = 1. 79 × 10 −^22 kg-m/sec

(e) The speed of the electron is most easily found from noting that

v c

pc E

= 0. 548 so v = 0. 548 c = 1. 64 × 108 m/sec

(f) Here is a momentum-energy diagram illustrating all of the above:

tan θ = pc E

= v c

= 0.

c^2 m^2 c^4

E (keV)

611 511

0 335 pc (keV)

θ

E^2 − p^2 =

K = 100

The moving electron is represented by the arrow, just as in the previous problem, with the arrow point on the hyperbola of mass-m particles. Note that the electron-volt units are much more convenient than our usual SI units, and lend greater understanding to the situation.