Converting ev/c^2 to joules

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.

Reply 1

uberteknik

21

Original post by 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.

m = ev/c²which is the form you are given on the question.......(eq 1)

You also know from Mr Einstein that

E = mc² therefore

m = E/c²..........(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^-19J and you have 494 million of the little blighters! :smile:

(edited 10 years ago)

Reply 2

RVNmax

19

Original post by uberteknik

m = ev/c²which is the form you are given on the question.......(eq 1)

You also know from Mr Einstein that

E = mc² therefore

m = E/c²..........(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^-18J and you have 494 million of the little blighters! :smile:

Original post by HerroKitty

Oh of course! Thank you! :smile:

But if you were to multiply

494MeV/c^2

by

1.6 \times 10^{19}

wouldn't you be getting units of

J/c^2

which is a mass rather than an energy.

That's why the OP was trying to multiply by

c^2

.
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/pages/physics_1/Particles/PP06/particles_6.htm

Please explain, I'm clearly missing something :redface:

(edited 10 years ago)

Reply 3

uberteknik

21

Original post by RVNmax

But if you were to multiply

494MeV/c^2

by

1.6 \times 10^{19}

wouldn't you be getting units of

J/c^2

which is a mass rather than an energy.

We are dealing with the equivalence between mass and energy. That is what E=mc² 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.
c² is a constant.

E = eV = Joules

m = E/c²= eV/c²

Reply 4

RVNmax

19

Original post by uberteknik

We are dealing with the equivalence between mass and energy. That is what E=mc² 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.
c² is a constant.

E = eV = Joules

m = E/c²= eV/c²

So basically, when I read 494eV/c², the eV/c² is not really the units, but a constant instead?

Reply 5

uberteknik

21

Original post by RVNmax

So basically, when I read 494eV/c², the eV/c² is not really the units, but a constant instead?

Nearly. c and hence 1/c² are the constants.

494MeV has the SI units of Joules. Where 1 eV is equivalent to 1.6x10^-19Joules

1/c² converts eV to the units of mass.

i.e. The units for c²cancel with the units of eV to leave the unit of mass on its own.
The units of mass multiplied with the units of c² equate to the units of eV.

The speed of light 'c' is the constant which cannot be exceeded. (c = 3x10⁸ms^-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 = d²s/dt² = ms^-2 therefore
Force = kg.ms^-2
Energy (Joules) = (kg.ms^-2).m = kg.m².s^-2

In the same way, eV = kg.m².s^-2 :

V = E/Q = JC^-1
Charge (Coulombs) = Current x Time = A.s
Voltage = kg.m².s^-2/ A.s
Voltage = kg.m².A^-1.s^-3

e = electron charge = A.s = 1.6x10^-19J
eV = A.s. kg.m².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:

c² = (ms^-1)² = m²s^-2
E = mc²
m = E/c² and substituting units
kg = kg.m².s^-2 / m².s^-2

kg = kg

The units equate.

(edited 10 years ago)

Reply 6

RVNmax

19

Original post by uberteknik

Nearly. c and hence c² are the constants.

It just seems weird that if only c² 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/c² = (494eV)/c² of which the units are eV/(ms^-1)²
Thus, E = 494eV/c² x c² = 494eV = 494 x 1.6 x 10^-19 J

(edited 10 years ago)

Reply 7

uberteknik

21