# Converting ev/c^2 to joules Watch

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I was wondering if someone could help me understand the conversion of MeV/C^2 into Joules, such as in the question below.

The mark scheme shows that the conversion is simply to multiply by 1.60x10^-13 to convert it into joules... but wouldn't the value of that be now J/c^2 instead of joules on its own?

Why do you not have to multiply the 494 by C^2 to get a value in MeV and then multiply by 1.60x10^-13 to convert to joules?

Thanks.

The mark scheme shows that the conversion is simply to multiply by 1.60x10^-13 to convert it into joules... but wouldn't the value of that be now J/c^2 instead of joules on its own?

Why do you not have to multiply the 494 by C^2 to get a value in MeV and then multiply by 1.60x10^-13 to convert to joules?

Thanks.

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#2

(Original post by

I was wondering if someone could help me understand the conversion of MeV/C^2 into Joules, such as in the question below.

The mark scheme shows that the conversion is simply to multiply by 1.60x10^-13 to convert it into joules... but wouldn't the value of that be now J/c^2 instead of joules on its own?

Why do you not have to multiply the 494 by C^2 to get a value in MeV and then multiply by 1.60x10^-13 to convert to joules?

Thanks.

**HerroKitty**)I was wondering if someone could help me understand the conversion of MeV/C^2 into Joules, such as in the question below.

The mark scheme shows that the conversion is simply to multiply by 1.60x10^-13 to convert it into joules... but wouldn't the value of that be now J/c^2 instead of joules on its own?

Why do you not have to multiply the 494 by C^2 to get a value in MeV and then multiply by 1.60x10^-13 to convert to joules?

Thanks.

^{2 }which is the form you are given on the question.......(eq 1)

You also know from Mr Einstein that

E = mc

^{2}therefore

m = E/c

^{2}..........(eq 2)

Comparing the two equations (eq 1 & 2) by inspection alone

E = ev !!!

The defintion of E for 1 electron volt is 1.6x10

^{-19}J and you have 494 million of the little blighters!

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#3

(Original post by

m = ev/c

You also know from Mr Einstein that

E = mc

m = E/c

Comparing the two equations (eq 1 & 2) by inspection alone

E = ev !!!

The defintion of E for 1 electron volt is 1.6x10

**uberteknik**)m = ev/c

^{2 }which is the form you are given on the question.......(eq 1)You also know from Mr Einstein that

E = mc

^{2}thereforem = E/c

^{2}..........(eq 2)Comparing the two equations (eq 1 & 2) by inspection alone

E = ev !!!

The defintion of E for 1 electron volt is 1.6x10

^{-18}J and you have 494 million of the little blighters!
(Original post by

Oh of course! Thank you!

**HerroKitty**)Oh of course! Thank you!

That's why the OP was trying to multiply by .

On this site, I think they've also done the same in question 3 and the explanation above it, where they've used kg in place of J.

http://www.antonine-education.co.uk/...articles_6.htm

Please explain, I'm clearly missing something

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#4

(Original post by

But if you were to multiply by wouldn't you be getting units of which is a mass rather than an energy.

**RVNmax**)But if you were to multiply by wouldn't you be getting units of which is a mass rather than an energy.

^{2}is saying. i.e. mass and energy are equivalent and interchangeable!

It's the fundamental law for all of physics.

When an atom decays, it releases energy by converting some of it's mass to energy. When a neutron decays in beta emission, it releases energy by converting some of its mass.

Energy IS defined as the electron volt.

c

^{2}is a constant.

E = eV = Joules

m = E/c

^{2 }= eV/c

^{2 }

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#5

(Original post by

We are dealing with the equivalence between mass and energy. That is what E=mc

It's the fundamental law for all of physics.

When an atom decays, it releases energy by converting some of it's mass to energy. When a neutron decays in beta emission, it releases energy by converting some of its mass.

Energy IS defined as the electron volt.

c

E = eV = Joules

m = E/c

**uberteknik**)We are dealing with the equivalence between mass and energy. That is what E=mc

^{2}is saying. i.e. mass and energy are equivalent and interchangeable!It's the fundamental law for all of physics.

When an atom decays, it releases energy by converting some of it's mass to energy. When a neutron decays in beta emission, it releases energy by converting some of its mass.

Energy IS defined as the electron volt.

c

^{2}is a constant.E = eV = Joules

m = E/c

^{2 }= eV/c^{2 }^{2 }, the eV/c

^{2 }is not really the units, but a constant instead?

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#6

(Original post by

So basically, when I read 494eV/c

**RVNmax**)So basically, when I read 494eV/c

^{2 }, the eV/c^{2 }is not really the units, but a constant instead?^{2}are the constants.

**494MeV has the SI units of Joules.**Where 1 eV is equivalent to 1.6x10

^{-19}Joules

**1/c**

^{2}converts eV to the units of mass.i.e. The units for c

^{2 }cancel with the units of eV to leave the unit of mass on its own.

The units of mass multiplied with the units of c

^{2}equate to the units of eV.

The speed of light 'c' is the constant which cannot be exceeded. (c = 3x10

^{8}ms

^{-1})

Looking at the base SI units by reducing and equating, we have:

**Energy:**

Energy = force x distance

Force = mass x acceleration, and

acceleration = dv/dt but

v = ds/dt then

acceleration = (ds/dt)dt = d

^{2}s/dt

^{2}= ms

^{-2}therefore

Force = kg.ms

^{-2}

**Energy**(Joules) = (kg.ms

^{-2}).m =

**kg.m**

In the same way,

^{2}.s^{-2}**eV**=

**kg.m**

^{2}.s^{-2}:**V**= E/Q =

**JC**

Charge (

^{-1}**Coulombs**) = Current x Time =

**A.s**

Voltage = kg.m

^{2}.s

^{-2 }/ A.s

**Voltage = kg.m**

^{2}.A^{-1}.s^{-3}e = electron charge = A.s = 1.6x10

^{-19}J

eV = A.s. kg.m

^{2}.A

^{-1}.s

^{-3}

eV = kg.m

^{2.}s

^{-2}

**electronvolt = kg.m**

^{2.}s^{-2}__So eV is also shown to have the units of Joules__

Finally:

**c**= (ms

^{2}^{-1})

^{2}=

**m**

^{2}s^{-2}E = mc

^{2}

m = E/c

^{2}and substituting units

kg = kg.m

^{2}.s

^{-2}/ m

^{2}.s

^{-2}

**kg = kg**

The units equate.

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#7

^{2}is the constant, as then you only have a portion of the full units of m being written.

So would it be acceptable to write m = E/c

^{2}= (494eV)/c

^{2}of which the units are eV/(ms

^{-1})

^{2}

Thus, E = 494eV/c

^{2}x c

^{2}= 494eV = 494 x 1.6 x 10

^{-19}J

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#8

(Original post by

So would it be acceptable to write m = E/c

Thus, E = 494eV/c

**RVNmax**)So would it be acceptable to write m = E/c

^{2}= (494eV)/c^{2}of which the units are eV/(ms^{-1})^{2}Thus, E = 494eV/c

^{2}x c^{2}= 494eV = 494 x 1.6 x 10^{-19}Jm = eV/c

^{2}

mc

^{2}= eVc

^{2}/c

^{2 }then

E = 494eV

E = 494 x 1.6x10

^{-19}J

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#9

(Original post by

Yes, because

m = eV/c

mc

E = 494eV

E = 494 x 1.6x10

**uberteknik**)Yes, because

m = eV/c

^{2}mc

^{2}= eVc^{2}/c^{2 }thenE = 494eV

E = 494 x 1.6x10

^{-19}J
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