Finding the de Broglie wavelength from momentum

[Matty R]

Homework Statement

What is the de Broglie wavelength of a neutron traveling with a momentum equal to 300 \frac{\text{MeV}}{\text{c}}?

Homework Equations

\lambda = \frac{h}{p}

The Attempt at a Solution

p = \frac{300 \cdot \left( \left(1\times10^6 \right) \times \left(1.602\times10^{-19} \right) \right)}{2.998\times10^8}

= 1.603\times10^{-19} \text{ kgms}^{-1}

\lamda = \frac{6.626\times10^{-34}}{1.603\times10^{-19}}

= 4.133\times10^{-15} \text{ m}

That's what I get, but the answer is given as 1.38x10^{-23}m.

By inserting this given answer into the equation, I get a value of 4.801x10^{-11} for p, which I can only get by the following:

300 \frac{\text{MeV}}{\text{c}} = 300 \cdot \left( \left( 1\times10^6 \right) \times \left(1.602\times10^{-19} \right) \right)

which ignores the c.

Is the given answer wrong, or am I missing something important?

This doesn't make any sense to me.

 

[Galileo's Ghost]

The unit MeV/c is actually a unit of momentum...no need to divide by the speed of light, it is already factored in through the use of this unit notation.

Generically, energy units consist of the quantities (mass)(length)² / (time)². If you were to divide energy by speed, you would be left with (mass)(length)/(time) which is a momentum unit.

 

[SammyS]

1 MeV/c ≈ 5.344285×10^-22 kg·m/s

So, 300 MeV/c ≈ 1.603×10^-19 kg·m/s as you calculated.

 

[Matty R]

Thanks for the replies.

Sorry. I'm still a bit confused.

To convert MeV/c to SI Units, do I just multiply the number (ie: 300) by "M" multiplied by "eV"?

I thought I was supposed to work it out as SammyS did, but doing that gives a different answer to the solutions.

EDIT

The unit MeV/c is actually a unit of momentum...no need to divide by the speed of light, it is already factored in through the use of this unit notation.

Generically, energy units consist of the quantities (mass)(length)² / (time)². If you were to divide energy by speed, you would be left with (mass)(length)/(time) which is a momentum unit.

Okay. MeV is a measurement of energy, units kgm^{2}s^{-2}. Dividing by speed (ms^{-1}) gives kgm^{-1}, which is momentum in SI units. So won't I need to divide my energy by speed, as you said?