There are two radioactive nuclei A and B . A is an alpha emitter and B is a beta emitter. Their disintegration constants are in the ratio of 1:2 . What should be the ratio of the number of atoms of A and B at any time t so that the probabilities of getting alpha and beta particles are the same at that instant?

Question

There are two radioactive nuclei A and B . A is an alpha emitter and B is a beta emitter. Their disintegration constants are in the ratio of 1:2 . What should be the ratio of the number of atoms of A and B at any time t so that the probabilities of getting alpha and beta particles are the same at that instant? (A) 2:1 (B) 1:2 (C) e (D) None

Tardigrade Admin · Accepted Answer

Given here, (λ A/λ B)=(1/2) Probability of getting α and β are same. Thus, rate of disintegration are equal. ∴ λ ANA=λ BNB=|(dN/dt)| Hence, the ratio of atoms is (NA/NB)=(λ B/λ A)=(2/1)

$2 : 1$

$1 : 2$

$e$

None

Given here, $\frac{λ _{A}}{λ _{B}} = \frac{1}{2}$
Probability of getting $α$ and $β$ are same.
Thus, rate of disintegration are equal.
$∴ λ_{A} N_{A} = λ_{B} N_{B} = ∣ ∣ \frac{d N}{d t} ∣ ∣$
Hence, the ratio of atoms is $\frac{N _{A}}{N _{B}} = \frac{λ _{B}}{λ _{A}} = \frac{2}{1}$

2:1

1:2

e

None

Given here, λB​λA​​=21​ Probability of getting α and β are same. Thus, rate of disintegration are equal. ∴λA​NA​=λB​NB​=∣∣​dtdN​∣∣​ Hence, the ratio of atoms is NB​NA​​=λA​λB​​=12​

$2 : 1$

$1 : 2$

$e$

Given here, $\frac{λ _{A}}{λ _{B}} = \frac{1}{2}$
Probability of getting $α$ and $β$ are same.
Thus, rate of disintegration are equal.
$∴ λ_{A} N_{A} = λ_{B} N_{B} = ∣ ∣ \frac{d N}{d t} ∣ ∣$
Hence, the ratio of atoms is $\frac{N _{A}}{N _{B}} = \frac{λ _{B}}{λ _{A}} = \frac{2}{1}$