Flag Modern Physics> A radioactive nucleus X decays to a nucle...
question mark

A radioactive nucleus X decays to a nucleus Y with a decay constant λx = 0.1 s-1 . Y further decays to a stable nucleus Z with a decay constant λy = 1/30 s-1. Initially, there are only X nuclei and their number is N0 = 1020. Set up the rate equations for the populations of X, Y and Z. The population of Y nucleus as a function or time is given by NY­ (t) = (N0 λλx – λy)) {exp (-λyt)-exp (-λxt) – exp (λx t)}. Find the time at which NY is maximum and determine the populations X and Z at that instant

Amit Saxena , 11 Years ago
Grade upto college level
anser 1 Answers
Navjyot Kalra
Hello Student,
Please find the answer to your question
The rate of equation for the population of X, Y and Z will be
dNx / dt = - λx Nx …(i)
dNy / dt = - λy Ny + λx Nx ….(ii)
dN / dt = - λy N­y …(iii)
⇒ On integration, we get
Nx = N0 / λx – λy [ e-λyt - eλxt]
To determine the maximum Ny, we find
dNy / dt = 0
From (ii)
- λyNy + λx Nx = 0
⇒ λx Nx = λyNy ………(v)
⇒ λx (N0 e-λxy) = λyx N0 / λx – λy (e-λyt - eλxt)]
⇒ λx – λy / λy = e-λyt – e-λxt / e-λxt ⇒ λx / λy = e(λx / λy)t
⇒ loge λx / λy = (λx – λy)t
⇒ t = logex / λy) / λx – λy = loge [0.1 / (1/30)] / 0.1 – 1/30 = 15 loge3
∴ Nx = N0 e-0.1(15 loge3) = N0 eloge(3-1.5)
⇒ Nx = N0 3-15 = 1020 / 3 √3
Since, dNy / dt = 0 at t = 15 loge 3, ∴ Ny = λxNx / λy = 1020 / √3
And N2 = N0 – Nx - Ny
= 1020 – (1020 / 3√3) – 1020 / √3 = 1020 (3 √3 – 4 / 3 √3)
Thanks
Navjot Kalra
askIITians Faculty
Last Activity: 11 Years ago
star
LIVE ONLINE CLASSES

Prepraring for the competition made easy just by live online class.

tv

Full Live Access

material

Study Material

removal

Live Doubts Solving

assignment

Daily Class Assignments