 GR 8677927796770177 | # Login | Register

GR9277 #85
Problem
 GREPhysics.NET Official Solution Alternate Solutions
\prob{85}
A free electron (rest mass 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 and the rest mass of the electron to be , one can figure out .

The momentum is given by .

Solve for the velocity from . Thus, the velocity is .

Plugging this into the equation for momentum, one gets , and thus its momentum is about 1.4, as in choice (C).  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= m c + (p c) then you already know that the momentum must be less than the energy, but greater than the difference between the energy and m c . This eliminates all but the correct answer. 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).
 Andresito2006-03-29 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.
 Andresito2006-03-29 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. EPdropout
2012-11-08 17:12:29
I used T=p/2m=(-1)mc, after finding =3
=>p=2m*(2/3)E
=>p=2.0 (MeV^2/c^2)
=>p =1.4 MeV/c rrfan
2011-11-06 17:41:09
A useful shortcut on these types of relativity problems:

The solution to is the following: .

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. dinoco
2010-11-07 07:52:37
You say that "v=2c/3." But since =8/9 then
v should be 2c/3 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= m c + (p c) then you already know that the momentum must be less than the energy, but greater than the difference between the energy and m c . This eliminates all but the correct answer.
 apr20102010-04-08 12:25:33 Remarkable
 thinkexist2012-10-12 09:29:49 This is actually so simple and brilliant I cannot believe I have not thought of this before.
 justin_l2013-10-15 23:50:31 if something is remarkably brilliant, would it not be more surprising that you *do* think of it? 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).
 Andresito2006-03-29 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.
 Andresito2006-03-29 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.
 ramparts2009-08-06 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.
 GREview2009-08-30 19:22:53 It may simplify things to not even think about the 's: Plugging and , we can solve for .
 Albert2009-11-05 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!      LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces . type this... to get...$\int_0^\infty\partial\Rightarrow\ddot{x},\dot{x}\sqrt{z}\langle my \rangle\left( abacadabra \right)_{me}\vec{E}\frac{a}{b}\$