Fusion of Deuterium Nuclei: Equation, Unknown Nucleus, and Energy Calculation

[fabbo]

Ok, I've been puzzling over this question - any thoughts?

A possible fusion process has deuterium nuclei combining to form an unknown nucleus plus one excess neutron. Write out an equation for this process. Identify the unknown nucleus. Use the following data to determine why deuterium is stable and calculate the energy released in the process. (Mass of proton=1.00783u, mass of neutron=1.00867u, mass of alpha particle=4.003u, mass of deuterium nuclide=2.01410u).

I have the unknown nucleus as Tritium. Not sure how to calculate energy released as I don't know the mass deficit. I thought about working out the mass deficit when an alpha particle is formed and multiplying that by 3/2 but not sure...

 

[fabbo]

The mass defect when an alpha particle formed is 0.03u. This is 27.945MeV and 1.75 x 1026J.

How can this be used to calculate the mass of the tritium formed?

 

[fabbo]

does anyone have any ideas?

 

[Astronuc]

The energy released in DT fusion is 17.6 MeV per reaction, which is

D + T -> n + $\alpha$

the mass defect is calculated as Q = (mass (reactants) - mass (products)) c², and using appropriate units.

1 MeV = 1.6022 x 10^-13 J.

Knowing the binding energy of the alpha particle will not help. What other information is given for the problem.

Is the mass defect of the alpha particle determined using the rest mass of two p's and 2 n's? If so, then all one can obtain is the binding energy per nucleon.

This is 27.945MeV and 1.75 x 1026J.

better check the conversion on MeV to J.

10²⁶ J is a lot of energy! A 1000 MWe plant produces 10⁹ J = 1 GJ of electrical energy in one second, or ~ 3.4 GJ of thermal energy in one sec.

 

[fabbo]

Knowing the binding energy of the alpha particle will not help. What other information is given for the problem.

.

there's no other information - that's why I am confused. I asked my teacher and he said it is solve - able...

 

[Astronuc]

Well, there is one equation - for mass defect, and two unknowns according to the problem as stated.

One would then need a second equation with the two unknowns, and offhand, I don't see that.

One could approximate tritium with the mass of d + n, but there is a difference of 0.007 amu.

The other matter is that I have different numbers for the proton, neutron, deuteron, and alpha particle.

m_p = 1.00727644 u
m_n = 1.00866452 u
m_d = 2.013553 u
m_a = 4.001503 u

and m_t = 3.015501 u

Looking at http://fusedweb.pppl.gov/CPEP/Chart_Pages/3.HowFusionWorks.html

m_p = 1.007276 u
m_n = 1.008665 u
m_d = 2.013553 u
m_a = 4.001506 u

and m_t = 3.015500 u

1 u = 1 atomic mass unit = 1.66054 x 10<sub>-27[/sup] kg = 931.466 MeV/c² (Ref PPPL)

These are close, but different than those provided in the OP.
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In order to use the other masses, one would have to know another Q value.

One could calculate Q of the reaction: p + n = d, and then try to approximate the Q for n + d = t, using the same mass defect for the ( p + n = d). But this still introduces a relatively large error.

However assuming this, then one can approximate (d + t) as (2d + n - Q(p + n = d)). But that would be a gross approximation.

The other part would be to use the momentum equation, but then one only had the momentum of the neutron equal (and opposite) the momentum of the alpha particle. The energy is partitioned according the masses of the neutron and alpha particle.

There are also DD reactions - d + d -> t + p and d + d -> He³ + n, but one would need Q values for these reactions.</sub>