Question

Calculate the kinetic energy of the ${ }^4 \mathrm{He}$ ion formed through thermal neutron capture in ${ }^{10} \mathrm{~B}$.

    Calculate the kinetic energy of the ${ }^4 \mathrm{He}$ ion formed through thermal neutron capture in ${ }^{10} \mathrm{~B}$.
Radiochemistry and Nuclear Chemistry
Radiochemistry and Nuclear Chemistry
Gregory Choppin,… 4th Edition
Chapter 11, Problem 7 ↓
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Calculate the kinetic energy of the ${ }^4 \mathrm{He}$ ion formed through thermal neutron capture in ${ }^{10} \mathrm{~B}$.
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Transcript

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00:01 So in the problem, we are told that there is a boron -neutron capture therapy that is conducted.
00:07 And this is the following nuclear process.
00:10 We are given the masses for each of the reactants and the products except for the proton.
00:15 And we are also given the total kinetic energy for the product.
00:19 And we are told that there is no kinetic energy release for the reactants.
00:23 And then we are ultimately asked to find what is the energy of the photon that is released.
00:29 So we're given the 2 .31 mega -electravolts, so we need to convert this into jewels.
00:37 Now, first off, we know that 2 .31 mega -electravolts is equal to 2 .31 times 10 to the 6 electro -volts.
00:49 Now, this comes in handy because now we can go from electrovolts to joules, because in 1, electrical volt is equal to 1 .60 to 18x10 .1 .8 times 10 to the negative 19th joules.
01:08 So using all this information, we can solve for, we can convert electrovolts to jewels.
01:14 So 2 .31 times 10 to the 6 electrovolts.
01:18 And then in one electrovolt, there's 1 .60218 times 10 to the negative 19 joules.
01:32 So this equals 3 .701 .035 times 10 to the negative 13 joules.
01:47 Now, since we have the energy, we can then solve for the mass by using, einstein's equation.
01:54 So when doing this we'll have to use e equals mc squared.
02:00 So let's rearrange this equation to solve for mass.
02:03 So this can be rearranged into mass equals energy divided by the speed of light squared.
02:13 So now we have to do this plug our value in that we just calculated into energy and then divide by the speed of light squared.
02:20 So when we do that, we will have the 3 .7015 times 10 to the negative 13 joules divide by the speed of light squared.
02:39 And this will give us our mass, and that is going to be 0 .4112 2 times 10 to negative 29 joules times second squared over meter squared.
02:59 Now, we must convert grams to dalton's.
03:05 So we know that joules times second over meters squared is going to be equal to one kilogram.
03:15 So these units equal one kilogram and one kilogram equals 1 ,000 grams.
03:27 So really what we have here is if we multiply this number times 1 ,000, we can get grams.
03:36 So really this is going to be 0 .41122 times 10 to the negative 26 grams.
03:48 So this is our mass right here because we have to get into units of grams for our next step when we have to convert grams to dalton's.
04:01 So for this, you must know that in one gram equal to one dalton over 1 .6605 times 10 to negative 24 grams.
04:18 So we can take our number of grams divided by this number to determine the number of dalton's.
04:23 So when doing that, 0 .41122 times 10 to the negative 26 grams, divided by 1 .6605 times 10 to the negative 24th grams, will give us our number of dalton's.
04:42 So that's going to be 0 .002476 dalton's.
04:50 So the whole reason why we have to convert grams to dalton's is because our masses are in dalton's, the ones that they gave us.
04:59 So this is a much simpler way rather than converting each one of these into grams.
05:04 So now that we have this in dalton's, we can find the mass of the proton, because we have our change in mass, which is going to be this number that we just calculated.
05:12 And now we can plug in the values that we're given for the products and reactants from the problem alone.
05:20 So let's do that.
05:22 So let's set this equal to 0.
05:25 0 .002476 dalton's equals, and now the sum of the products, so we have 4 .002603 plus 7 .011604 plus, and this is our unknown, this is what we're interested in solving...
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