Simple Atomic Mass Unit question

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Homework Help Overview

The discussion revolves around a nuclear reaction involving the uranium isotope 235U capturing a neutron and undergoing fission to produce rubidium-93 and cesium-141. Participants are exploring the calculation of energy released in this process, referencing nuclear masses and the relationship between mass and energy.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of nuclear masses and the energy released based on mass differences. Questions arise regarding how the nuclear masses are determined and the significance of the neutron's mass in the calculations. There is also inquiry into the reasoning behind the inclusion of two neutrons in the reaction equation.

Discussion Status

The discussion is active, with participants providing insights into the mass-energy relationship and the necessity of using measured values for nuclear masses. Some participants are questioning the assumptions made about the reaction and the conservation of nucleon number, indicating a productive exploration of the topic.

Contextual Notes

Participants note that the masses of the isotopes must be taken from a table, and there is an emphasis on the importance of the neutron's mass in the calculations. The discussion also highlights the need for clarity on the total nucleon number in the reaction.

ZedCar
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Homework Statement


The uranium isotope 235U captures a neutron and undergoes fission to produce 93Rb and 141Cs. Calculate the energy released in this process.

The nuclear masses of the relevant isotopes are
235U 235.0439u,
93Rb 92.9217u,
141Cs 140.9195u



Homework Equations





The Attempt at a Solution



The answer is given as

n + 235U → 93Rb + 141Cs + 2n

LHS 235.0439 + 1.0087
RHS 92.921712 + 140.91949 + 2.10087

Difference is 0.1943u = 181 MeV


I realize the quantity of 1.0087 is the mass of neutron divided by one unified atomic mass unit. So that's (1.67493 x 10^-27) / (1.66 x 10^-27).
Also, the 2.10087 number is simply double this.

How have the 235.0439, 92.921712 & 140.91949 values been calculated? I know it's to do with E=Mc^2

Thank you.
 
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It all has to do with the mass difference, (mass before) - (mass after) = (released) or (consumed energy). The values for the nuclei´s masses has to be taken from a table. There is no way of calculating the mass of a nuclei, it has to be measured.
 
Can it not be calculated from the part of the question which states;

The nuclear masses of the relevant isotopes are
235U 235.0439u,
93Rb 92.9217u,
141Cs 140.9195u
 
The difference in mass is what is relevant, ((Mass before)-(Mass after)).*c^2 = E. To calculate this you also need the mass of the neutron. The mass of the neutron can not be calculated and needs to be taken from a table. Also in order to get the mass in kilograms for the other nuclei you simply multiply it with the atomic mass unit, just as with the neutron.
 
In the answer, how is it known that 2n should be added? Why 2?
 
ZedCar said:
In the answer, how is it known that 2n should be added? Why 2?

Take a look at the reaction - how many neutrons are produced?
 
Is it because the total nucleon number on both sides must equal. Since LHS=236 RHS can only equal this by adding 2n.
 

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