. co An αparticle ( ^4 He. nucleus ) is to be taken apart in the following steps. Give the energ

. co An αparticle ( ^4 He. nucleus ) is to be taken apart in...

$.$ co An $\alpha$ particle $\left({ }^{4} \mathrm{He}\right.$ nucleus $)$ is to be taken apart in the following steps. Give the energy (work) required for each step: (a) remove a proton, (b) remove a neutron, and (c) separate the remaining proton and neutron. For an $\alpha$ particle, what are (d) the total binding energy and (e) the binding energy per nucleon? (f) Does either match an answer to (a), (b), or (c)? Here are some atomic masses and the neutron mass. $\begin{array}{llll}{ }^{4} \mathrm{He} & 4.00260 \mathrm{u} & { }^{2} \mathrm{H} & 2.01410 \mathrm{u} \\ { }^{3} \mathrm{H} & 3.01605 \mathrm{u} & { }^{1} \mathrm{H} & 1.00783 \mathrm{u} \\ \mathrm{n} & 1.00867 \mathrm{u} & & \end{array}$

$.$ co An $\alpha$ particle $\left({ }^{4} \mathrm{He}\right.$ nucleus $)$ is to be taken apart in the following steps. Give the energy (work) required for each step: (a) remove a proton, (b) remove a neutron, and (c) separate the remaining proton and neutron. For an $\alpha$ particle, what are (d) the total binding energy and (e) the binding energy per nucleon? (f) Does either match an answer to (a), (b), or (c)? Here are some atomic masses and the neutron mass.
$\begin{array}{llll}{ }^{4} \mathrm{He} & 4.00260 \mathrm{u} & { }^{2} \mathrm{H} & 2.01410 \mathrm{u} \\ { }^{3} \mathrm{H} & 3.01605 \mathrm{u} & { }^{1} \mathrm{H} & 1.00783 \mathrm{u} \\ \mathrm{n} & 1.00867 \mathrm{u} & & \end{array}$

Key Concepts

Alpha Particle Structure

An alpha particle, essentially a helium-4 nucleus, consists of two protons and two neutrons tightly bound together. Its study provides insight into the fundamental interactions within a small nucleus and serves as an important example in discussions about nuclear stability and the distribution of binding energy among nucleons.

Mass–Energy Equivalence

Mass–energy equivalence, encapsulated in Einstein’s equation E=mc², is the principle that mass can be converted into energy and vice versa. This formulation is key to nuclear physics as it explains how a small loss of mass in a nucleus can result in a significant amount of binding energy, thereby explaining energy release in nuclear processes.

Nuclear Binding Energy

Nuclear binding energy represents the energy required to disassemble a nucleus into its individual protons and neutrons. It is a measure of the stability of a nucleus; the larger the binding energy, the more stable the nucleus. This concept is fundamental in understanding why energy is released in nuclear reactions and how the forces within the nucleus operate.

Mass Defect

The mass defect is the difference between the total mass of the individual nucleons and the actual mass of the nucleus. This discrepancy arises because some of the mass is converted into binding energy during the formation of the nucleus according to the mass–energy equivalence. It is this mass loss that accounts for the energy holding the nucleus together.

Separation Energy

Separation energy is the energy required to remove a single nucleon or a cluster of nucleons from a nucleus. It varies depending on the specific nucleon and its interactions with the remainder of the nucleus. This concept is essential for understanding the steps in breaking apart a nucleus into its constituent particles.

Transcript

00:01 In this problem on the topic of nuclear physics, we are told that an alpha particle, which is a helium nucleus, is to be taken apart with various steps.

00:08 We want to find the energy or the work required for each step.

00:12 Firstly, by removing a proton, then removing a neutron, then separating the remaining proton and neutron.

00:19 And then we want to know for an alpha particle, what is the total binding energy, the binding energy per nucleon.

00:24 And we want to know if our answers here match either a, b or c.

00:31 Now, the first step is to add energy to produce the helium nucleus h .e .4 becomes a proton plus h3, which, to make the electron balance, may be written as he4 and a hydrogen nucleus.

01:01 H1 plus h3.

01:06 And the energy needed for this, we'll call it delta e1, is equal to the mass of h3 plus the mass of the proton minus the mass of the alpha particle all times c squared.

01:28 And so this is 3 .01605 u.

01:36 Plus the mass of proton 1 .007 -8 -3u minus 4 .002 -6 u times the energy equivalent of c squared, which is 931 .5 mega -electron volts per atomic mass unit u.

02:02 And so this becomes 19 .8 mega electron volts to remove the proton.

02:21 For the second step, to add energy produces h3, which then becomes a neutron plus h2.

02:36 And so the energy needed delta e2 for this process is equal to the mass of the of h2 plus the mass of the neutron minus the mass of h3 times c squared...

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