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Boron Neutron-Capture Therapy In boron neutron-capture therapy (BNCT), a patient is given a compound containing $^{10} \mathrm{B}$ that accumulates inside cancer tumors. Then the tumors are irradiated with neutrons, which are absorbed by $^{10} \mathrm{B}$ nuclei. The product of neutron capture is an unstable form of $^{11} \mathrm{B}$ that undergoes $\alpha$ decay to $^{7} \mathrm{Li}$. a. Write a balanced nuclear equation for the neutron absorption and $\alpha$ decay process. b. Calculate the energy released by each nucleus of boron- 10 that captures a neutron and undergoes $\alpha$ decay, given the following masses of the particles in the process: $^{10} \mathrm{B}(10.0129 \mathrm{amu}),^{7} \mathrm{Li}(7.01600$ amu) $^{4} \mathrm{He}(4.00260 \mathrm{amu}),$ and $^{1} \mathrm{n}(1.00866 \mathrm{amu})$ c. Why is the formation of a nuclide that undergoes $\alpha$ decay a particularly effective cancer therapy?

   Boron Neutron-Capture Therapy In boron neutron-capture therapy (BNCT), a patient is given a compound containing $^{10} \mathrm{B}$ that accumulates inside cancer tumors. Then the tumors are irradiated with neutrons, which are absorbed by $^{10} \mathrm{B}$ nuclei. The product of neutron capture is an unstable
form of $^{11} \mathrm{B}$ that undergoes $\alpha$ decay to $^{7} \mathrm{Li}$.
a. Write a balanced nuclear equation for the neutron absorption and $\alpha$ decay process.
b. Calculate the energy released by each nucleus of boron- 10 that captures a neutron and undergoes $\alpha$ decay, given the following masses of the particles in the process: $^{10} \mathrm{B}(10.0129 \mathrm{amu}),^{7} \mathrm{Li}(7.01600$ amu)
$^{4} \mathrm{He}(4.00260 \mathrm{amu}),$ and $^{1} \mathrm{n}(1.00866 \mathrm{amu})$
c. Why is the formation of a nuclide that undergoes $\alpha$ decay a particularly effective cancer therapy?

Chemistry The Science in Context

Thomas R. Gilbert 5th Edition

Chapter 19, Problem 69

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The equation starts with the boron-10 nucleus capturing a neutron to form an unstable boron-11 nucleus. This unstable nucleus then undergoes alpha decay to form a lithium-7 nucleus and an alpha particle (helium-4 nucleus). The balanced nuclear equation Show more…

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00:06 Boron neutron capture is a type of cancer therapy where patients are given a compound containing boron 10 which accumulates inside cancer tumors.

00:22 This is then the cancer tumor is then irradiated with neutrons.

00:29 So these neutrons can get absorbed by the boron 10 nuclei and form an unstable form of boronon which is a boron 11.

00:39 So this is unstable and will undergo alpha decay to form stable lithium 7 nuclei.

00:50 So let's try to write down the balanced nuclear reaction or equation associated with this process that i just described.

01:03 So boron 10 has the atomic number 5.

01:11 It can capture a nuclear nuclear reaction.

01:14 Neutron and form boron 11.

01:23 That will then undergo alpha decay and form an alpha particle along with a lithium nucleus.

01:44 So we can also get rid of the boron 11 in the middle and have this as a single nuclear reaction.

01:57 Next, let's try to calculate the energy released by the energy.

02:01 A single boron 10 nucleus when it captures a neutron and undergoes alpha decay.

02:09 So we know the masses of all of these nuclei.

02:16 You know boron 10, the mass for a neutron, the mass for an alpha particle, and the mass for a lithium 7 nuclei.

02:28 So when two nucleon, when two nucleon, or a nuclei with a nucleon in this case combines to form a different nuclei.

02:46 There is always a mass difference here.

02:50 So the masses here is always going to be less than the mass of the reactants.

03:07 So let's say mass reactants, mass product.

03:13 So the masses of the reactants, reactant nuclei, the sum of the masses is always going to be greater than the masses of the product.

03:23 So there's a mass difference there.

03:26 So we can calculate that mass difference and then use the einstein equation to determine the energy released in this reaction.

03:35 So we are given the masses of each nuclear in this case.

03:41 And all we had to do was sum the masses of the two reactants and the masses of the two products...

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