# Calculate the temperature for this reaction to occur

1. May 13, 2016

### Janiceleong26

1. The problem statement, all variables and given/known data

2. Relevant equations
KE = 3/2 R/NA T
3. The attempt at a solution
By using the formula above, I used 2.4 x 10^-14 J as the mean translational KE , but it should be 1.2x10^-14 , why? I thought the mixture contains both deuterium nucleus and the proton? Why half KE of mixture?

2. May 13, 2016

### drvrm

what happens to the particles kept at some temperature-they get energy and start thermal motion - the average K.E. of particles is related to the temp. and Boltzmann Constant.
3/2 is a factor depending on degree of freedom.
so, what amount of energy a particle say proton will get if you are given with Total K.E. of proton and deuterium? that one should find and then apply the energy-temp relation.

3. May 13, 2016

### haruspex

At a given temperature, how much KE will each particle have? (You have to assume some collisions will be head on.)

4. May 13, 2016

### Janiceleong26

Half of the total KE, but why? I thought they are asking for the sample? Why do we need to consider KE of just one of them?

5. May 13, 2016

### haruspex

They are asking for the temperature at which you would expect the reaction to occur. It doesn't have to occur for all particle pairs immediately. Leaving aside that some will have greater than the average KE, they will be moving in various directions. From the temperature you can find the average KE of any given particle, but the reaction involves two particles. Their relative directions of travel matter. Which case makes them most likely to reach the required KE level?

6. May 13, 2016

### drvrm

i think the average velocity of a particle in an enclosure at Temp. T may be taken as v(rms) and
1/2 . m(i). v(i)^2 = 3/2 k T ; now one has to check how the total KE will be shared.

7. May 14, 2016

### haruspex

Not sure what you mean by that. Shared between what?

8. May 14, 2016

### drvrm

shared between the two kinds of particles namely deuteron and protons.

9. May 14, 2016

### haruspex

10. May 14, 2016

### drvrm

well it does -but the question is being asked as to why half of the total KE of the two (given in the question) is being used for estimating T.

11. May 14, 2016

### Janiceleong26

Does it mean that we have to supply 1.2x10^-14 J of energy to both deuterium and the proton for the reaction to start?

12. May 14, 2016

### haruspex

It means that given those average KEs each, you can expect the reaction to start.

13. May 14, 2016

### Janiceleong26

Ok, got it, thanks very much!