Neutron, proton collision problem

[ofeyrpf]

The following problem appeared on the A2 Edexcel Physics unit 4 exam paper January 2012 question 18. The solution, as given by the exam board, is attached.

Question:
18. James Chadwick is credited with discovering the neutron in 1932.

Beryllium was bombarded with alpha particles, knocking neutrons out of the beryllium atoms. Chadwick placed various targets between the beryllium and a detector. Hydrogen and nitrogen atoms were knocked out of the targets by the neutrons and the kinetic energies of these atoms were measured by the detector.

(a) The maximum energy of a nitrogen atom wa found to be 1.2 MeV.

Show that the maximum velocity of the atom is about 4 x 10⁶ m/s.

mass of nitrogen atom = 14u, where u = 1.66 x 10^-27 kg

Solution:
The set up as I understand it is,
alpha --> Be --> neutron --> target --> Ni or H --> detector

v = sqrt(2(1.2x10^-6x1.6x10^-19)/(14x1.66x10^-19)) = 4.06x10^-6 m/s No problems here.

Question
(b)The mass of a neutron is Nu (where N is the relative mass of the neutron) and its initial velocity is x. the nitrogen atom, mass 14u, is initially stationary and is then knocked out of the target with a velocity, y, by a collision with a neutron.

(i) Show that the velocity, z, of the neutron after the collision can be written as

z = (Nx - 14y)/N​

Solution:
momentum before = momentum after

Nux = 14uy - Nuz

rearranging gives,

z = (Nx - 14y)/N No problems here.

Question
(ii)The collision between this neutron and the nitrogen atom is elastic. What is meant by an elastic collision?

Solution
In an elastic collision the kinetic energy is conserved. No problems here.

Question
(iii) Explain why the kinetic energy E_k of the nitrogen atom is given by

E_k = (Nu(x² - z²)/(2)​

Solution:
Using conservation of kinetic energy,

E_K(n) = E_k(Ni) + E_k(n)

(1/2)Nux² = (1/2)14Nuy² = (1/2)Nuz²

y² = (x² - y²)/(14)

E_k(Ni) = (1/2)14Nuy²

= (1/2)(14Nu(x² - z²)/14)

= (Nu(x² - z²)/(2)

For this calculation to work, the mass of the Ni has to be 14Nu but in the question it is given as 14u. That is the first thing I don't understand.

Question
(c) The two equations in (b) can be combined and z can be eliminated to give

y = (2Nx)/(N + 14)​

Solution
The question does not ask how this is done but I'd like to know and can't figure it out. I tried substituting

z = (Nx - 14y)/(N) into E_k = (Nu(x² - z²))/(2)

and this gives,

(2E_k)/(Nu) = x² - ((Nx - 14y)/(N))²

But this has an E_k in it, so I don't see how to get to the required y = (2nx)/(N + 14) This is the second problem I have, not understanding where this equation comes from.

Question
(i) The maximum velocity of hydrogen atoms knocked out by neutrons in the same experiment was 30 x 10⁷ m/s. The mass of a hydrogen atom is 1u.

Show that the relative mass N of the neutron is 1.

Solution
There is an error in the question here. Instead of 30 x 10⁷ m/s it should 3.0 x 10⁷ m/s.

The equation given in the question applies to Nitrogen and can be rearranged to give

2Nx = y_Ni(N + 14) = 4.1 x 10⁶_Ni(N + 14)

y_Ni = 4.1 x 10⁶ m/s. This is obtained from part (a).

For hydrogen then

2nx = y_H(N + 1) = 3.0 x 10⁷(N+1)

These two equation can be combined giving,

4.1 x 10⁶_Ni(N + 14) = 3.0 x 10⁷(N+1)

from which N can be solved

N = (3 x 10⁷-14 x 4.1 x 10⁶)/(4.1 x 10⁶ - 3 x 10⁷)
= 1.05 which is approximately 1

Question
(ii) This equation can not be applied to all collisions in this experiment. Suggest why.

Solution
As the atoms approach the speed of light their mass does not remain constant, it increases.

 

[haruspex]

(1/2)Nux² = (1/2)14Nuy² = (1/2)Nuz²

Shouldn't that read:
(1/2)Nux² = (1/2)14uy² + (1/2)Nuz²
?

 

[ofeyrpf]

Neutron, Ni, H, collision problem

Hi haruspex,

Thanks for your reply. Looking back on this post I can't believe how many typos I've made.

Yes you are correct it should read...

18. (b) (iii)

E_K(n) = E_k(Ni) + E_k(n)

(1/2)Nux² = (1/2)14uy² + (1/2)Nuz²

14uy² = Nu(x² - z²)

y² = N(x² - z²)/14

E_k(Ni) = (1/2)14uy²

= (1/2)(14uN(x² - z²)/14)

= Nu(x² - z²)/(2)

So that is correct then, thanks.

Now if I could just figure out where they get the equation in part (c), I'd be happy.

 

[haruspex]

For c, you have two equations:
Nx²=14y²+Nz²; Nz = Nx - 14y
Just eliminate z between them.

 

[ofeyrpf]

$$
Nx^2=14y^2+Nz^2\\
\mbox{and}\\
Nz=Nx-14y\\
\mbox{so,}\\
z^2=\frac{(Nx-14y)(Nx-14y)}{N^2}\\
=\frac{N^2x^2-28yNx+196y^2}{N^2}\\
\mbox{substituting this into the first equation gives,}\\
Nx^2=14y^2+N\left(\frac{N^2x^2-28yNx+196y^2}{N^2}\right)\\
0=Ny^2-2yNx+14y^2\\
2Nx=y(N+14)\\
y=\frac{2Nx}{n+14}
$$
Thanks for your help,

Shane

 

[ofeyrpf]

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