There are two radio nuclei A and B. A is a $α$ -emitter and B is $β$ -emitter. Their disintegration constant are in the ratio of 1:2. What should be the number of atoms ratio of A and B at time t=0, so that initially probability of getting of $α$ and $β$ -particles are same.

Open in App

Solution

$\frac{λ_{A}}{λ_{B}} = \frac{1}{2}$
Probability of getting alpha and beta particles are same $.$
Thus rate of disintegration are equal
$λ_{A} N_{A} = λ_{B} N_{B}$
$\frac{λ_{A}}{λ_{B}} = \frac{λ_{A}}{λ_{B}} = 1 : 2$
Hence,
we get.
$\frac{λ_{A}}{λ_{B}} = \frac{λ_{A}}{λ_{B}} = 1 : 2$

Suggest Corrections

Similar questions

Q. There are two radio nuclei A and B. A is an

α

-emitter and B a

β

- emitter. Their disintegration constants are in the ratio of

1 : 2

. What should be 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?

Q. There are two radio nuclei A and B. A is an alpha emitter and B a beta emitter. Their disintegration constants are in the ratio of 1 : 2. What should be 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?

Q. Two radioactive substances X and Y emit

α

and

β

particles respectively. Their disintegration constants are in the ratio 2:3. To have equal probabilities of getting emission of

α

and

β

particles, find the ratio of number of atoms of X to that of Y at any time instant.

Q. Two Radioactive substances X and Y emit

α

and

β

particles respectively. Their disintegration constants are in the ratio 2 : 3. To have equal rate of disintegration of getting emission of

α

and

β

particles., the ratio of number of atoms of X to that of Y at any time instant is

Two Radioactive substances X and Y emit $α$ and $β$ particles respectively. Their disintegration constants are in the ratio 2 : 3. To have equal rate of disintegration of getting emission of $α$ and $β$ particles., the ratio of number of atoms of X to that of Y at any time instant is

Join BYJU'S Learning Program

There are two radio nuclei A and B. A is a α-emitter and B is β-emitter. Their disintegration constant are in the ratio of 1:2. What should be the number of atoms ratio of A and B at time t=0, so that initially probability of getting of α and β-particles are same.

λAλB=12Probability of getting alpha and beta particles are same.Thus rate of disintegration are equal λANA=λBNBλAλB=λAλB=1:2Hence,we get.λAλB=λAλB=1:2

There are two radio nuclei A and B. A is a $α$ -emitter and B is $β$ -emitter. Their disintegration constant are in the ratio of 1:2. What should be the number of atoms ratio of A and B at time t=0, so that initially probability of getting of $α$ and $β$ -particles are same.

$\frac{λ_{A}}{λ_{B}} = \frac{1}{2}$
Probability of getting alpha and beta particles are same $.$
Thus rate of disintegration are equal
$λ_{A} N_{A} = λ_{B} N_{B}$
$\frac{λ_{A}}{λ_{B}} = \frac{λ_{A}}{λ_{B}} = 1 : 2$
Hence,
we get.
$\frac{λ_{A}}{λ_{B}} = \frac{λ_{A}}{λ_{B}} = 1 : 2$