Miscalculation with speeds of electrons?

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In summary, to increase the speed of an electron from 1.2*10^8 m/s to 2.4*10^8 m/s, 0.294 MeV of work needs to be done.
  • #1
asdf1
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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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  • #2
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:
  • #3
i think I'm missing something...
[tex]m_ec^2 = .511 Mev[/tex]
i thought that [tex]m_e[/tex]=0.511?
 
  • #4
[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.)
 
  • #5
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
 
  • #6
my math is crummy...
um, isn't units and the numbers multiplied separately?
so m=(0.511*c^2) MeV?
 
  • #7
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:
  • #8
thank you! :)
 

1. What is a miscalculation with speeds of electrons?

A miscalculation with speeds of electrons refers to an incorrect estimation or measurement of the velocity at which electrons are moving within a given system.

2. What causes miscalculations with speeds of electrons?

Miscalculations with speeds of electrons can be caused by a variety of factors such as measurement errors, limitations of equipment, or incorrect assumptions about the system being studied.

3. How can miscalculations with speeds of electrons impact scientific research?

Miscalculations with speeds of electrons can lead to inaccurate data and conclusions, which can impact the validity and reliability of scientific research. It can also hinder our understanding of the behavior and properties of electrons.

4. How can scientists avoid miscalculations with speeds of electrons?

To avoid miscalculations with speeds of electrons, scientists must use precise and accurate measurement techniques, carefully consider the limitations of their equipment, and constantly evaluate and revise their assumptions about the system being studied.

5. Are there any real-world applications for understanding miscalculations with speeds of electrons?

Yes, understanding miscalculations with speeds of electrons is crucial in fields such as materials science, electronics, and quantum physics. It can help improve the design and performance of electronic devices, as well as advance our understanding of fundamental principles in physics.

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