Miscalculation with speeds of electrons?

1. Oct 9, 2005

asdf1

for the following question:
how much work (in MeV) must be done to increase the speed of an electron from 1.2*10^8 m/s to 2.4*10^8 m/s?

my problem:
E= (gamma)mc^2=m(c^2){1/[1-(2.4/30^2]-1/[1-(1.2/30^2]}
=0.511(c^2)[1/(0.6)-1/(0.84)^(1/2)]=2.65*10^16

the correct answer should be 0.294 MeV~

does anybody know what went wrong?

2. Oct 9, 2005

Andrew Mason

$$\Delta E = (\gamma_2 - \gamma_1)m_ec^2$$

$$\gamma_1 = (1-v_1^2/c^2)^{-1/2} = 1.091$$
$$\gamma_2 = (1-v_2^2/c^2)^{-1/2} = 1.667$$
$$m_ec^2 = .511 Mev$$

$$\Delta E = .576 * .511 = .294 MeV$$

AM

Last edited: Oct 9, 2005
3. Oct 9, 2005

asdf1

i think i'm missing something...
$$m_ec^2 = .511 Mev$$
i thought that $$m_e$$=0.511?

4. Oct 9, 2005

Staff: Mentor

$m_e$ is the mass of the electron: $9.11 \ 10^{-31}$ kg. If you calculate $m_e c^2$ in standard units, you'll get the answer in Joules. Then convert Joules to eV. (1 eV = $1.60 \ 10^{-19}$ J.)

5. Oct 9, 2005

Andrew Mason

When mass is written in terms of an energy, it is understood that it is in units of Energy/c^2. The 1/c^2 is often omitted when it is written, so it can be confusing. So, $m_e = .511 Mev/c^2$ and $m_ec^2 = .511 MeV$.

AM

6. Oct 10, 2005

asdf1

my math is crummy...
um, isn't units and the numbers multiplied separately?
so m=(0.511*c^2) MeV?

7. Oct 10, 2005

Andrew Mason

Well, the units are really MeV/(9e16 m^2/sec^2) which works out to 1.78e-30 kilograms. But kilograms is not a very useful unit when measuring the mass of an electron. So we just use units of MeV/c^2 or MeV-mass

$$m \ne .511 c^2 MeV$$

$$m = .511 (MeV/c^2) units = .511 MeV(mass)$$
$$= .511e6/9e16 eV/m^2/sec^2 = .511e6/9e16 *1.6e(-19) kg$$

AM

Last edited: Oct 10, 2005
8. Oct 10, 2005

asdf1

thank you!!! :)