# 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 λλ (λx – 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

Navjyot Kalra
askIITians Faculty 654 Points
9 years ago
Hello Student,
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