Purchase Solution

Exploring the decay of Plutonium 241

Not what you're looking for?

Ask Custom Question

Pu-241 is produced from neutron absorption by Pu-240 (which is produced by neutron absorption by Pu-239). Calculate the fractional in-growth of Am-241 from Pu-241 (i.e. number ratio Am-241/Pu-241) from time = 0 to 200 years. Pu-241 is the parent and Am-241 is the daughter.

Purchase this Solution

Solution Summary

The decay of Plutonium 241 to Americium 241 is explored through its decay equations and the fractional growth in am-241 from Pu 241 is determined

Solution Preview

The amount of 241Am in a plutonium sample arises from it's formation in the decay of 241Pu and it's destruction from it's own decay. The differential equation governing the number of atoms of 241Am in a sample is

dN(A) / dt = -Lamda(A)*N(A) + K(P)*Lamda(P)*N(P) (1)

where

N(A) = Number of atoms of 241Am at time t

N(P) = Number of atoms of 241Pu at time t

Lamda(A) = Decay constant of 241Am

Lamda(P) = Decay constant of 241Pu

Lamda(n) = rate at which 240Pu absorbs neutrons to form 241Pu

N(240) = Number of atoms of 240Pu in the original sample

K(P) = 0.9999754, fraction of 241Pu decays that lead to 241Am, accounting for 2.4 x 10-3 % that decay to 237U #

{We will assume this to be K(P) = 1 for the rest of the ...

Purchase this Solution


Free BrainMass Quizzes
The Moon

Test your knowledge of moon phases and movement.

Intro to the Physics Waves

Some short-answer questions involving the basic vocabulary of string, sound, and water waves.

Variables in Science Experiments

How well do you understand variables? Test your knowledge of independent (manipulated), dependent (responding), and controlled variables with this 10 question quiz.

Introduction to Nanotechnology/Nanomaterials

This quiz is for any area of science. Test yourself to see what knowledge of nanotechnology you have. This content will also make you familiar with basic concepts of nanotechnology.

Classical Mechanics

This quiz is designed to test and improve your knowledge on Classical Mechanics.