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  GR9277 #85
Problem
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\prob{85}
A free electron (rest mass $m_e-0.5MeV/c^2$ has a total energy of 1.5MeV. Its momentum p in units of MeV/c is about

  1. 0.86
  2. 1.0
  3. 1.4
  4. 1.5
  5. 2.0

Special Relativity}Momentum

Given a total energy of \gamma mc^2 = 1.5MeV and the rest mass of the electron to be m_e = .5MeV/c^2, one can figure out \gamma =3.

The momentum is given by p=\gamma mv = 3mv=3v/2 (MeV/c^2).

Solve for the velocity from \gamma= \frac{1}{\sqrt{1-\beta^2}} \Rightarrow \gamma^2 = \frac{1}{1-\beta^2} \Rightarrow 1 = \gamma^2(1-\beta^2) \Rightarrow \beta^2 = 1-\gamma^{-2} \Rightarrow \beta^2=8/9. Thus, the velocity is v=2c/3.

Plugging this into the equation for momentum, one gets p=3/2\times \sqrt{8/9}=\sqrt{2}, and thus its momentum is about 1.4, as in choice (C).

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
unoriginal5279
2007-04-11 09:24:43
When short on time (as I always am) I try to avoid calculations whenever possible. If you remember that E\c^2= m\c^2 c\c^4 + (p c)\c^2 then you already know that the momentum must be less than the energy, but greater than the difference between the energy and m c\c^2 . This eliminates all but the correct answer.Alternate Solution - Unverified
Andresito
2006-03-29 15:15:31
You could always use the general equation,

E^2 = (p*c)^2 + (m*c^2)^2

Solve for p having E = 3/2, and m*c^2 = 1/2

to obtain 1.4 as in (C).
Andresito
2006-03-29 15:21:28
The equation I describe is beter to use since you only ned to take a square root once, and that helps when having larger numbers, see in test 9277 problem 70.
Andresito
2006-03-29 15:24:03
The equation I describe is beter to use since you only ned to take a square root once, and that helps when having larger numbers, see in test 9277 problem 70.
Alternate Solution - Unverified
Comments
EPdropout
2012-11-08 17:12:29
I used T=p\^2/2m=(\gamma-1)mc\^2, after finding \gamma=3
=>p\^2=2m*(2/3)E
=>p\^2=2.0 (MeV^2/c^2)
=>p =1.4 MeV/c
NEC
rrfan
2011-11-06 17:41:09
A useful shortcut on these types of relativity problems:

The solution to \gamma=\frac{1}{\sqrt{1-\beta^2} is the following: \beta=\sqrt{1-\frac{1}{\gamma^2}.

Even if you don't memorize this, it is straightforward to derive and is often useful on multiple problems, saving some time on the algebra.
NEC
dinoco
2010-11-07 07:52:37
You say that "v=2c/3." But since \beta^2=8/9 then
v should be 2\sqrt{2}c/3
Typo Alert!
unoriginal5279
2007-04-11 09:24:43
When short on time (as I always am) I try to avoid calculations whenever possible. If you remember that E\c^2= m\c^2 c\c^4 + (p c)\c^2 then you already know that the momentum must be less than the energy, but greater than the difference between the energy and m c\c^2 . This eliminates all but the correct answer.
apr2010
2010-04-08 12:25:33
Remarkable
thinkexist
2012-10-12 09:29:49
This is actually so simple and brilliant I cannot believe I have not thought of this before.
justin_l
2013-10-15 23:50:31
if something is remarkably brilliant, would it not be more surprising that you *do* think of it?
Alternate Solution - Unverified
Andresito
2006-03-29 15:15:31
You could always use the general equation,

E^2 = (p*c)^2 + (m*c^2)^2

Solve for p having E = 3/2, and m*c^2 = 1/2

to obtain 1.4 as in (C).
Andresito
2006-03-29 15:21:28
The equation I describe is beter to use since you only ned to take a square root once, and that helps when having larger numbers, see in test 9277 problem 70.
Andresito
2006-03-29 15:24:03
The equation I describe is beter to use since you only ned to take a square root once, and that helps when having larger numbers, see in test 9277 problem 70.
ramparts
2009-08-06 22:54:12
Well, it's better to use in problem 70, but I think it's pretty clear that for this question, the E^2 equation takes a lot less calculation than the "official" answer.
GREview
2009-08-30 19:22:53
It may simplify things to not even think about the c's:

E = \sqrt{p^2+m^2}

Plugging E=3/2 and m=1/2, we can solve for p.
Albert
2009-11-05 00:32:16
Hey, I see what you did there, you used the c's in the denominator of the units right along side the values, smart work. Best solution!
Alternate Solution - Unverified

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