# Homework Help: Uncalculable v

1. Oct 8, 2005

### asdf1

for the following question:
find the momentum of an electron whose kinetic energy equals its rest energy of 511 keV.

my problem:
[(gamma)-1]mc^2=511*1000*1.6*10^(-19)
but when i try to calculate the v in the gamma factor, it's impossible to calculate on the calculator because the number is too small...
any suggestions?

2. Oct 8, 2005

### Tide

I'm not sure I see where your difficulty is. It appears that $\gamma = 2$ so $\beta = \frac {\sqrt 3}{2}$. Did I miss something?

3. Oct 8, 2005

### asdf1

@@a
i think i'm the one who's missing something... why's gamma=2? i thought that you were supposed to try the "v" in gamma (gamma= [1-v^2/(c^2)]^(-1)) so isn't gamma unknown? may i also ask what does beta represent?

4. Oct 8, 2005

### Physics Monkey

The energy of the electron is $$E = \gamma m_0 c^2 = K + m_0 c^2$$ where $$K$$ is the kinetic energy of the electron. You are told that $$K = m_0 c^2$$, right? Solve for $$\gamma$$.

5. Oct 8, 2005

### Tide

You have $\gamma - 1 = 1$ so that $\gamma = 2$. Also, $\beta$ is just v/c.

6. Oct 9, 2005

### asdf1

wow... i hadn't thought of that way to solve the problem...
so because my way is almost impossible to solve, so you guys changed your method to using the formula
$$E = \gamma m_0 c^2 = K + m_0 c^2$$?

7. Oct 9, 2005

### lightgrav

Doing it your way, use m=0.911e-30 kg, c=3e8 m/s
to get gamma = 2.0 ... what did *you* use for m?

beta is v/c (comes before gamma, inside the gamma formula)

8. Oct 10, 2005

### asdf1

i used m=9.1*10^(-31) kg

9. Oct 11, 2005

### asdf1

is that method right?