Question

a radioactive substance disintegrates into two types of daughter nuclei, one type with disintegration constant λ1 and the other type with disintegration constant λ2, determine the half life of the radioactive substance​

Answer 1

Answer:

Half life of the radioactive substance is given as

$T_{1/2} = \frac{ln 2}{\lambda_1 + \lambda_2}$

Explanation:

As we know that the substance decay into two different nuclei

So rate of decay of Nuclei X will be equal to the rate of formation of its two daughter nuclei

So we have

$\lambda_{eq} N_o = \lambda_1 N_o + \lambda_2 N_o$

so we have

$\lambda_{eq} = \lambda_1 + \lambda_2$

here we know that half life is given as

$T_{1/2} = \frac{ln2}{\lambda}$

so we have

$\frac{ln 2}{T_{1/2}} = \lambda_1 + \lambda_2$

so we have

$T_{1/2} = \frac{ln 2}{\lambda_1 + \lambda_2}$

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Topic : Radioactivity

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Answer 2

Given :

A radioactive substance disintegrate in two daughter nuclei having disintegration constants

λ1 and λ2.

To find :

Half life of parent substance.

Solution :

We know that on disintegration of parent substance,

two substances are forming.

Lets name the parent substance A and daughter nuclie B and C with disintegration constants λ1 and λ2.

so,

Rate of disappearance of A = Rate of appearance of B + Rate of appearance of C.

(equation 1)

we know the formula that,

$\frac{dN}{dt} = - \lambda \: N$

(equation 2)

where LHS is rate of disappearance of a radioactive substance.

and λ is the disintegration constant and N = number of that substance particles.

from equation 1,

$\bf- \frac{dN_A }{dt} = \frac{dN_B \: }{dt} + \frac{dN_C \: }{dt}$

(equation 3)

Rate of appearance of B :

$\bf \frac{dN_B \: }{dt} = \lambda_1 \times N_A \:$

Rate of appearance of C :

$\bf \frac{dN_C \: }{dt} = \lambda_2 \times N_A \: \:$

Putting these valus in equation 3 , we get

$\bf - \frac{dN_ A\: }{dt} = \lambda_1 \times N_A + \lambda_2 \times N_A \: \: \\ \bf - \frac{dN_ A\: }{dt} = (\lambda_1 + \lambda_2) \times N_A$

On comparing with equation 1,

Effective disintegration constant

λ = λ1 + λ2.

We know that

Half life of a radioactive substance is :

$\bf \: t _{1/2} = \frac{0.693}{ \lambda}$

Putting the value of effective disintegration constant,

Half life of this radioactive substance :

$\bf \: t _{1/2} = \frac{0.693}{ (\lambda_1 + \lambda_2)}$