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Solved: Nuclear Fission. The unstable nucleus of
Chapter 23, Problem 87P(choose chapter or problem)
Problem 87P
Nuclear Fission. The unstable nucleus of uranium 236 can be regarded as a uniformly charged sphere of charge \(Q=+92 e\) and radius \(R=7.4 \times 10^{-15}\) m. In nuclear fission, this can divide into two smaller nuclei, each with half the charge and half the volume of the original uranium-236 nucleus. This is one of the reactions that occurred in the nuclear weapon that exploded over Hiroshima, Japan, in August 1945. (a) Find the radii of the two “daughter” nuclei of charge \(+46 e\). (b) In a simple model for the fission process, immediately after the uranium-236 nucleus has undergone fission, the “daughter” nuclei are at rest and just touching, as shown in Fig. P23.87. Calculate the kinetic energy that each of the “daughter” nuclei will have when they are very far apart. (c) In this model the sum of the kinetic energies of the two “daughter” nuclei, calculated in part (b), is the energy released by the fission of one uranium-236 nucleus. Calculate the energy released by the fission of 10.0 kg of uranium-236. The atomic mass of uranium-236 is 236 u, where 1 \(u=1\) atomic mass unit \(=1.66 \times 10^{-24}\) kg. Express your answer both in joules and in kilotons of TNT (1 kiloton of TNT releases \(4.18 \times 10^{12}\) J when it explodes). (d) In terms of this model, discuss why an atomic bomb could just as well be called an “electric bomb.”
Equation Transcription:
Text Transcription:
Q=+92e
R=7.4x10^-15
+46e
u=1
=1.66x10^-24
4.18x10^12
Q=+92e
Q=+46e
Q=+46e
Questions & Answers
QUESTION:
Problem 87P
Nuclear Fission. The unstable nucleus of uranium 236 can be regarded as a uniformly charged sphere of charge \(Q=+92 e\) and radius \(R=7.4 \times 10^{-15}\) m. In nuclear fission, this can divide into two smaller nuclei, each with half the charge and half the volume of the original uranium-236 nucleus. This is one of the reactions that occurred in the nuclear weapon that exploded over Hiroshima, Japan, in August 1945. (a) Find the radii of the two “daughter” nuclei of charge \(+46 e\). (b) In a simple model for the fission process, immediately after the uranium-236 nucleus has undergone fission, the “daughter” nuclei are at rest and just touching, as shown in Fig. P23.87. Calculate the kinetic energy that each of the “daughter” nuclei will have when they are very far apart. (c) In this model the sum of the kinetic energies of the two “daughter” nuclei, calculated in part (b), is the energy released by the fission of one uranium-236 nucleus. Calculate the energy released by the fission of 10.0 kg of uranium-236. The atomic mass of uranium-236 is 236 u, where 1 \(u=1\) atomic mass unit \(=1.66 \times 10^{-24}\) kg. Express your answer both in joules and in kilotons of TNT (1 kiloton of TNT releases \(4.18 \times 10^{12}\) J when it explodes). (d) In terms of this model, discuss why an atomic bomb could just as well be called an “electric bomb.”
Equation Transcription:
Text Transcription:
Q=+92e
R=7.4x10^-15
+46e
u=1
=1.66x10^-24
4.18x10^12
Q=+92e
Q=+46e
Q=+46e
ANSWER:
Introduction
First we have to calculate the radius of the daughter nucleus. Then we have to calculate the kinetic energy of the daughter nucleus when they are very far from each other using the conservation of energy. Then we have to calculate the energy released by 10 kg uranium in fission. Then we have to discuss why an atomic bomb can also be called as an electric bomb.
Step 1
(a) The total volume of the nucleus will remain same.
The initial radius of the nucleus was
Hence the initial volume is
So the volume of the each daughter nucleus is
If the radius of the daughter nucleus is , then the volume is
Hence we have
m.