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Chapter 19: Problem 126

There are two radioactive nuclei \(A\) and \(B . A\) is an alpha emitter and \(B\) is a beta emitter. If their disintegration constants are in the ratio \(1: 2\), then the ratio of number of atoms of \(A\) and \(B\) at any time \(t\) so that probabilities of getting alpha and beta particles are same at that instant

Short Answer

Expert verified
The ratio of the number of atoms of A to B at any time t such that the probabilities of getting alpha and beta particles are the same is 2:1

Step by step solution

01

Establish the relationship

The number of decaying atoms depends on the product of the volume of the sample, the number of atoms per unit volume, and the probability per unit time that an atom will decay. Hence the probability of decay is \(\lambda N\). For each radioactive nuclide A and B respectively, the probabilities are \(\lambda_A N_A\) and \(\lambda_B N_B\).
02

Form an equation based on given conditions

We are given that the probabilities of emitting alpha and beta particles are equal, which means the probabilities of decay for A and B are equal. Thus, we have \(\lambda_A N_A = \(\lambda_B N_B\).
03

Plug in the value of disintegration constants

We are given that the disintegration constants \(\lambda_A\) and \(\lambda_B\) are in the ratio 1:2. Let's denote \(\lambda_A\) as \(\lambda\) for simplicity, then \(\lambda_B\) becomes 2\(\lambda\). Replace these values into the equation, we have \(\lambda N_A = 2\(\lambda\) N_B\). The \(\lambda\) cancels out on both sides.
04

Find the ratio

Isolate N_B on one side to find the ratio of N_A to N_B: \(N_A / N_B = 2\), which means the ratio of the number of atoms of A to B is 2:1

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