Answer to Question #216881 in Civil and Environmental Engineering for syra

Question #216881
Calculate the nuclear binding energy ( in J) and the nuclear binding energy per nucleon of 209/83 Bi (208.9804 amu)
1
Expert's answer
2021-07-16T03:56:26-0400

1. The nucleus has 83 protons and 126 neutrons that is 209 nucleons total. With the known data of the mass of the atom  (1.007825 AMU) and neutron (1.008665 AMU), it is possible to run the following analysis:

The mass of 83 atoms of  (that is, the mass of 83 protons and 83 electrons) is,

"83 * 1.007825 amu = 83.649475 amu"

and the mass of 126 neutrons is,

"126 * 1.008665 amu = 127.09179 amu"

Mass of reagents:

"83.649475 amu + 127.09179 amu = 210.741265 amu"

Together exceed the measured mass for. The difference between the mass of an atom and the sum of the masses of the protons, neutrons, and electrons is called the mass defect. The missing mass can be explained by the theory of relativity, which states that mass loss is the energy released to the surroundings. According to the relation of mass-energy equivalence developed by Einstein (E = mc2), it is possible to calculate the amount of energy released.

"\u0394E = (\u0394m)c^2"

where;

ΔE = energy products - energy reagent

Δm = mass product - mass reagents

c = speed of light ="2.998*10^8 m\/s"

so;

Δm = 208.9804 amu - 210.741265 amu

= -1.760865 amu


As it has less mass than calculated from electrons and nucleons present, Δm is negative. Therefore, ΔE also is a negative amount; ΔE is released to the surroundings after forming the nucleus. So ΔE is calculated as follows:

"\u0394E = -1.760865 amu (2.998*10^8 m \/ s) ^2\n\n= -1.58*10^{17} amu m^2 \/ s^2"

applying conversion factors;

"1 kg = 6.022*1026 amu\\\\\n\n1 J = 1 kg.m^2 \/ s^2"

is obtained,

"\u0394E = (-1.58*10^{17} uma m^2 \/ s^2) (1 kg \/ 6.022*10^{26} amu) (1 J \/ 1kg.m2 \/ s2)\n\n= -2.68*10^{-10} J"

This is the amount of energy released when the Bi-83 core is formed from 83 protons and 126 electrons. The nuclear binding energy is "-2.68*10^{-10} J"  and represents the amount of energy needed to break the nucleus into protons and neutrons individual.


2. However, when the stability of two cores compared with anyone should consider having a different number of nucleons; for this reason, it makes more sense to use the nuclear binding energy per nucleon, which is defined as:

nuclear binding energy per nucleon = nuclear binding energy/nucleon number

For the core Bi-83,

nuclear binding energy per nucleon ="2.68x10-\\frac{10 J}{ 209} nucleons\n\n= 1.25x10-12 J \/ nucleons"

The nuclear binding energy per nucleon is 1.25x10-12 J / nucleons and compares the stability of all cores with a common base.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS