Fusion reaction (don't know what I'm doing wrong)

[tkoi]

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

Although in most problems you solve in this course you should use values of constants rounded to 2 or 3 significant figures, in this problem you must keep at least 5 significant figures throughout your calculation. Problems involving mass changes require many significant figures because the changes in mass are small compared to the total mass. In this problem you must use the following values of constants, accurate to 5 significant figures:

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 1.38e-13 joules, and the proton starts out with a kinetic energy of 2.77e-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. What will be the total kinetic energy of both particles an instant before they touch?
K_{1_H} + K_{2_H} = 2.86826e-13 joules (I got this one right)

What is the kinetic energy of the reaction products (helium nucleus plus photon)?
K_{He} + K_{\gamma} = ________ joules

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

For the first question, which I got right:

E_f = E_i + W

E_f = E_i + U_i - U_f

E_f + U_f = E_i + U_i

K_{1,f} + K_{2,f} + m_1 c^2 + m_2 c^2 + U_f = K_{1,i} + K_{2,i} + m_1 c^2 + m_2 c^2 + U_i

K_{1,f} + K_{2,f} + U_f = K_{1,i} + K_{2,i} + U_i
The particles start out 'far apart' so U_i = 0

K_{1,f} + K_{2,f} = K_{1,i} + K_{2,i} - U_f
= 2.77 \times 10^{-13} + 1.38 \times 10^{-13} - \frac{1}{4 \pi \epsilon_0} \times \frac{e \times e}{2 \times 0.9 \times 10^{-15}}
= 2.86826 \times 10^{-13}


The second question:

E_f + U_f = E_i + U_i

K_3 + K_\gamma + m_3 c^2 + m_\gamma c^2 + U_f = K_2 + K_1 + m_2 c^2 + m_1 c^2 + U_i
Gamma ray has 0 mass and 0 charge, so

K_3 + K_\gamma + m_3 c^2 = K_2 + K_1 + m_2 c^2 + m_1 c^2 + U_i

K_3 + K_\gamma = K_2 + K_1 + m_2 c^2 + m_1 c^2 - m_3 c^2 + U_i

= K_2 + K_1 + c^2 (m_2 + m_1 - m_3) + \frac{1}{4 \pi \epsilon_0} \times \frac{e \times e}{r}

= 1.38 \times 10^{-13} + 2.77 \times 10^{-13} + c^2 (m_2 + m_1 - m_3) + \frac{1}{4 \pi \epsilon_0} \times \frac{e \times e}{2 \times 0.9 \times 10^{-15}}
Here remember that m_n = (rest mass) * (atomic mass unit)

= 1.34905 \times 10^{-12}
My homework is submitted online, and this answer is wrong. I don't know where I messed up, I did exactly what I did for the first question which I got right. Any help is appreciated.

 

[willem2]

You added the potential energy when the proton and deuteron nearly touch to the kinetic energy when they are still far apart to get the initial energy.

 

[tkoi]

Ah! Of course it was something that simple. Sometimes I guess you just need a fresh pair of eyes to see what went wrong. Thanks a lot!