Miscalculation with speeds of electrons?

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Homework Help Overview

The discussion revolves around calculating the work required to increase the speed of an electron from 1.2 x 10^8 m/s to 2.4 x 10^8 m/s, focusing on relativistic energy concepts and the mass-energy equivalence principle.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the calculation of energy changes using relativistic gamma factors and question the interpretation of mass in energy units. There is a discussion on the correct application of formulas and the conversion between units of energy and mass.

Discussion Status

The discussion includes attempts to clarify misunderstandings about the mass-energy equivalence and the proper use of units in calculations. Some participants provide insights into the conversion of mass to energy and the implications of using different units.

Contextual Notes

Participants express confusion regarding the relationship between mass and energy, particularly in the context of relativistic physics. There is mention of the need to convert between Joules and electronvolts, as well as the significance of using the correct units for mass when applying the equations.

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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?
 
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asdf1 said:
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

[tex]\Delta E = (\gamma_2 - \gamma_1)m_ec^2[/tex]

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

[tex]\Delta E = .576 * .511 = .294 MeV[/tex]

AM
 
Last edited:
i think I'm missing something...
[tex]m_ec^2 = .511 Mev[/tex]
i thought that [tex]m_e[/tex]=0.511?
 
[itex]m_e[/itex] is the mass of the electron: [itex]9.11 \ 10^{-31}[/itex] kg. If you calculate [itex]m_e c^2[/itex] in standard units, you'll get the answer in Joules. Then convert Joules to eV. (1 eV = [itex]1.60 \ 10^{-19}[/itex] J.)
 
asdf1 said:
i think I'm missing something...
[tex]m_ec^2 = .511 Mev[/tex]
i thought that [tex]m_e[/tex]=0.511?
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, [itex]m_e = .511 Mev/c^2[/itex] and [itex]m_ec^2 = .511 MeV[/itex].

AM
 
my math is crummy...
um, isn't units and the numbers multiplied separately?
so m=(0.511*c^2) MeV?
 
asdf1 said:
my math is crummy...
um, isn't units and the numbers multiplied separately?
so m=(0.511*c^2) MeV?
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

[tex]m \ne .511 c^2 MeV[/tex]

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

AM
 
Last edited:
thank you! :)
 

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