# Atomic scale unit conversions

1. Mar 19, 2006

### 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…

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

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…

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

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

Where did I go wrong?

2. Mar 19, 2006

### 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:

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

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 wont leave in in terms of some number divided by c. You will carry out the divison and then give your solution.

Last edited: Mar 19, 2006
3. Mar 19, 2006

### 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.

4. Mar 19, 2006

### 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.

5. Mar 19, 2006

### Cyrus

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

6. Mar 19, 2006

### 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

7. Mar 20, 2006

### 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.

8. Mar 20, 2006

### Pengwuino

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

9. Mar 21, 2006

### 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.