# How do I solve for the wavelength of an alpha particle?

1. Nov 24, 2015

### KP2727

1. The problem statement, all variables and given/known data
Compare 25-MeV alpha particles wavelengths with the size of a nucleon in a nucleus.

2. Relevant equations
λ = h/p
E2 = p2c2 + m2c4
KE = ½mv2
p = mv

3. The attempt at a solution
To compare the wavelength of an α particle with a nucleon, I would need to divide the wavelength of this 25-MeV α particle by the size of a nucleon, 1×10-15 (I looked this value up). So I'm looking for λα / dnucleon.

First I solved for p using the second relevant equation, which gave me:
p = √((E2 - (mc2)2)/c2).

E = 25 MeV, or the energy of the alpha particle, and
mc2 = 4.00150618 u⋅c2 × 931.5 MeV/c2 = 3727.403 MeV. So
p = √(((25 MeV)2 - (3727.403 MeV)2)/c2). This, however, gives me a negative number under the square root.

I then tried to use the third and fourth relevant equations to get p. The mass of the α particle in kilograms is 6.64465675×10-27. I converted the α particle's energy from MeV to joules:
25 MeV = 4.0054 × 10-12 J = 4.0054 × 10-12 kg⋅m2/s2

Solving for v gave me:
v = √(2(4.0054×10-12 kg⋅m2/s2) / 6.64465675×10-27 kg)
v = 34721755.06 m/s

p = mv
p = (6.64465675×10-27 kg)(34721755.06 m/s) = 2.30714114×10-19

Plugging in p for the wavelength formula gives me:
λ = h/p = (6.626×10-34 kg⋅m2/s) / (2.30714114×10-19 kg⋅m/s)
λ = 2.87195223×10-15 m

Finally, I divide λ by the size of a nucleon, which would be 1×10-15.
λα / dnucleon = 2.87195223×10-15 m / 1×10-15 m
λα / dnucleon = 2.87195

However, this answer is incorrect. The actual answer at two significant figures is 1.2, and I can't for the life of me figure out what I did wrong. I have a second problem exactly like this one, but with a proton particle instead, so I'm hoping that figuring out what I did wrong on this problem will help me solve the second problem. If anyone can help me, it would be much appreciated!

Last edited: Nov 24, 2015
2. Nov 24, 2015

### TSny

Regarding your first attempt, the energy given in the problem is the kinetic energy of the alpha particle, not the total energy E.

Your second method looks OK to me. The "size" of a nucleon could refer to its diameter. You might check a few other web sites for values of the radius of a proton or neutron. The value depends on how the radius is defined (or measured).

3. Nov 24, 2015

### KP2727

Assuming my second method in finding λα is correct, that would mean dnucleon would have to be 2.39329×10-15 m. I've looked at multiple websites on the sizes of nucleons, protons, and neutrons, and just about all of them say that it's 1×10-15 m. Some sites say the diameter is 0.8418 fm or 2×10-14 m, but dividing λα by either of these numbers would result in a number even farther from the actual answer.

So unless the question has a different nucleon size in mind, I feel like the mistake is somewhere in my calculations.

4. Nov 24, 2015

### TSny

A couple of things:

The "size of a nucleon" might refer to the diameter of the nucleon, rather than the radius
EDIT: See http://phys.org/news/2013-11-proton-radius-puzzle-quantum-gravity.html for accepted radius of an isolated proton. Or, https://en.wikipedia.org/wiki/Proton

The problem asks you to compare the wavelength with the size of a nucleon in a nucleus. The effective size of a nucleon when inside a nucleus could be different than the size of an isolated nucleon. It might be helpful to consider the approximate formula for the radius, R, of a nucleus in terms of the number of nucleons, A. See the formula for R here: https://en.wikipedia.org/wiki/Atomic_nucleus#Nuclear_models You can use this to deduce the effective radius of a nucleon inside of a nucleus to about 2 significant figures.

Last edited: Nov 24, 2015
5. Nov 24, 2015

### KP2727

I think that formula helped me solved my problem! Every time I searched for the size of a nucleon, every result said that the average diameter of a nucleon is 1 fm, which is why I kept using that value even after you said "The size of a nucleon might refer to the diameter of the nucleon, rather than the radius."

This time, I used r0 from the R formula you provided as the nucleon radius (1.25×10-15 m), and doubled that to get the diameter, or the nucleon size (2.5×10-15 m). I divided λα by that, and got 1.14878, which is much closer to the actual answer.

To make sure that was the issue, I solved the second problem I mentioned I had with the proton particle using 2r0 as the nucleon size again, and I got very close to the actual answer (off by about 0.1, but likely due to simple rounding errors).

I can't thank you enough for your help. I spent hours on this problem unable to figure out where I messed up!

6. Nov 24, 2015

### TSny

OK. Good. Note, the value of $r_o$ varies in the literature. But I've often seen it given as 1.2 fm

7. Jan 10, 2017

### lkrustev

Guys is it possible to make a double slit experiment with alpha particles?