Converting Atomic Units: Where Did I Go Wrong?

[Pengwuino]

I basically am having a problem understanding the units used on the atomic level such as MeV/c and MeV/c^2.

I have a problem where I have a given energy of 370 MeV for a photon.

This means the wavelength is

\lambda = \frac{{(6.6261*10^{ - 34} )c}}{{(3.7*10^8 eV*\frac{{1.6022*10^{ - 19} J}}{{1eV}})}}

\lambda = 3.35*10^{ - 15} m

The photons momentum is also given by…

\begin{array}{l}<br /> p = \frac{{6.6261*10^{ - 34} }}{{3.35*10^{ - 15} m}} = \frac{h}{\lambda } \\ <br /> p = (1.978*10^{ - 19} kg*m/s)(\frac{{1\frac{{MeV}}{c}}}{{5.344*10^{ - 22} \frac{{kg*m}}{s}}}) = 370\frac{{MeV}}{c} \\ <br /> \end{array}

This begs the question, did I screw up and get into some circular logic or is the energy in MeV/c^2 the same number as the momentum is in MeV/c for photons?

Also, in these units, I need to determine what an equivalent particle's would be with that total energy and momentum. I used…

\begin{array}{l}<br /> E^2 = p^2 c^2 + m^2 c^4 \\ <br /> m^2 c^4 = E^2 - p^2 c^2 \\ <br /> m^2 = \frac{{E^2 }}{{c^4 }} - \frac{{p^2 }}{{c^2 }} \\ <br /> m = \sqrt {370^2 \frac{{MeV^2 }}{{c^4 }} - 370^2 \frac{{MeV^2 }}{{c^4 }}} \\ <br /> \end{array}

Obviously I did something wrong… Doesn't c = 1 somewhere?

Where did I go wrong?

 

[Cyrus]

Where did this:

\lambda = \frac{{(6.6261*10^{ - 34} )c}}{{(3.7*10^8 eV*\frac{{1.6022*10^{ - 19} J}}{{1eV}})}}

come from? You should show your steps to make it clear.

E = \frac {hc}{\lambda}

h= 6.6260(52) 10^{-34} J*s

c = 3.0x10^8

307Mev = 370(10^6)*1.602(10^{-19})

\lambda = 3.353(10^{-15})

You look good so far.

This:

\begin{array}{l} p = \frac{{6.6261*10^{ - 34} }}{{3.35*10^{ - 15} m}} = \frac{h}{\lambda } \\ p = (1.978*10^{ - 19} kg*m/s)

Seems ok too, now, to answer your question:

This begs the question, did I screw up and get into some circular logic or is the energy in MeV/c^2 the same number as the momentum is in MeV/c for photons?

I don't see MeV/c^2 anywhere. Why are you dividing by c^2?

It seems to me that from the equation:

p = \frac{E}{c} that yes, the value of momentum will have the same value as the energy, divided by the constant c. But that is quite meaningless, as c is a number that should be equal to 3.0(10^8). It is not a unit, it is assigned a value. So you when you find an answer you won't leave in in terms of some number divided by c. You will carry out the divison and then give your solution.

 

[Meir Achuz]

To make things simple, forget about the c and hbar.
Use the conversion 1=197 MeV-fm, just as you use
1=2.54 cm/inch.
Then lambda for a 370 Mev photon is 197/370=0.43 fm.

If E=p, the mass must be zero.

 

[Pengwuino]

oops, i guess the real problem I'm looking at is to why the momentum's # equal to the energy's # (and not it's real world meaning)? Is that a coincadence or is it a consequence of the formulas for calculating energy of a photon and it's momentum using the units of MeV and MeV/c and MeV/c^2, etc etc.

When I look at E = \frac {hc}{\lambda} and p = \frac {h}{\lambda}, it appears like the same number comes up to be the same as a consequence of the formulas used.

What's actually confusing me is how to figure out the mass of an equivalent particle. I can see how I can turn E into the correct units since it obviously divides by c^2 but the momentum is tricky. I divide by c^2 and get the correct units but it all becomes 0 which can't be the mass.

 

[Cyrus]

I can't find this, MeV/c^2, where did you get this from?

 

[Pengwuino]

m^2 = \frac{{E^2 }}{{c^4 }} - \frac{{p^2 }}{{c^2 }}

\frac{{E^2 }}{{c^4 }}

That'll come out to be \frac{{MeV^2 }}{{c^4 }}

Square root it to determine the mass and you have the units of mass of MeV/c^2

 

[Meir Achuz]

Just forget abot the c and hbar. HE physicists have been doing that for 50 years, except when they write books. E and p are components of the same four-vector. Just use MeV for each, and there is no problem.
m^2=E^2-p^2, if all are in MeV. You can convert from Mev to 1/fm, if you want to, by using 1=197 MeV-fm.

 

[Pengwuino]

But the p^2 is in MeV/c^2 and energy is in MeV^2

 

[Meir Achuz]

Just never write in the c and you can't go wrong.
p can be given in MeV since it is in the same four-vector as E.
c is just a conversion factor, used if you happen to use different units for different parts of the same vector.
It would be the same if you insisted on using cm for x and y,
but feet for z.