Fusion Reaction of Hydrogen and Deuterium into Helium

[ankitgu]

Homework Statement

In a fusion reaction, the nuclei of two atoms join to form a single atom of a different element. In such a reaction, a fraction of the rest energy of the original atoms is converted to kinetic energy of the reaction products. A fusion reaction that occurs in the Sun converts hydrogen to helium. Since electrons are not involved in the reaction, we focus on the nuclei.
Hydrogen and deuterium (heavy hydrogen) can react to form helium plus a high-energy photon called a gamma ray:

$$^1H + ^2H \rightarrow ^3He + \gamma$$

Objects involved in the reaction:

Particle        # of protons    # of neutrons    Charge    Rest Mass (atomic mass units)
[sup]1[/sup]H (proton)     1               0                +e        1.0073
[sup]2[/sup]H (deuterium)  1               1                +e        2.0136
[sup]3[/sup]He (helium)    2               1                +2e       3.0155
gamma ray       0               0                0         0

Constant                  Value to 5 significant figures
c (speed of light)        2.9979e8 m/s
e (charge of a proton)    1.6022e-19 coulomb
atomic mass unit          1.6605e-27 kg
[tex]\frac{1}{4 \pi \epsilon_0}[/tex]                      8.9875e9 N·m[sup]2[/sup] /C[sup]2[/sup]

A proton (1H nucleus) and a deuteron (2H nucleus) start out far apart. An experimental apparatus shoots them toward each other (with equal and opposite momenta). If they get close enough to make actual contact with each other, they can react to form a helium-3 nucleus and a gamma ray (a high energy photon, which has kinetic energy but zero rest energy). Consider the system containing all particles. Work out the answers to the following questions on paper, using symbols (algebra), before plugging numbers into your calculator.

The deuterium nucleus starts out with a kinetic energy of 1.54e-13 joules, and the proton starts out with a kinetic energy of 3.08e-13 joules. The radius of a proton is 0.9e-15 m; assume that if the particles touch, the distance between their centers will be twice that.

1. What will be the total kinetic energy of both particles an instant before they touch? (I got this)

2. What is the kinetic energy of the reaction products (helium nucleus plus photon)? (I need help)

Homework Equations

$$E_f = E_i + W$$
$$W = -\Delta U = U_i - U_f$$
$$E = m c^2 + K$$
$$U_{elec} = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r}$$

The Attempt at a Solution

Number 1:
Answer: 3.34e-13
DeltaE = DeltaKE + Delta PE
0 = KEf-KEi+PEf-PEi
KEf = what we want
KEi = given
PEf = 9e9*(-1.6e-19)^2/1.8e-15
PEi = 0

Number 2:
DeltaE = KEf - KEi + PEf - PEi
0 = KEf - KEi + PE(coloumbs final) - PE(coloumbs initial) + PE(R final) - PE(R initial)
KEf = what we want
KEi = answer to #1
PE(coloumbs final) = 0, b/c photon has no mass
PE(coloumbs initial) = 9E9*(-1.6e-19)^2/1.8E-15
PE(R final) = Mass(helium)*c^2
PE(R initial) = Mass(deuterium + proton)*c^2
My answer for KEf = 2.71395E17

Where did I go wrong on #2?

 

[gneill]

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