Question

Two radioactive substances A and B have deacy constants 5 ohn and 1ohm respectively at t = 0 they have the same number of nuclie.The ratio of number of nuclie of A to those of B will be [1/e]^2 after a time interval​

Answer 1

Q] ___?

=> 4 λ !

Explanation:

Number of nuclei remained after time can be written as

N = No e^λt

Where, No is initial number of nuclei of both the substances,

N1 = Noe^ -5λt _[i]

N2 = Noe^- λt _[ii]

Dividing Eq. [i] by Eq [ii], we obtain

N1 / N2 = e^( -5λt + λ) t = e-4λt = 1/e^4λt

But, we have given

N1 / N2 = [ 1/e ]^2 = 1/e^2

Hence, 1/e^2 = 1/e-4λt

Comparing the powers, we get

2 = 4 λt

or t = 2 / 4λt = 1 /2λ

The interval will be 4 λt !

Answer 2

Answer:

Answer :

Number of nuclei remained after time can be written as

• N = No e^λt

Where, No is initial number of nuclei of both the substances,

• N1 = Noe^ -5λt _[i]
• N2 = Noe^- λt _[ii]

Dividing Eq. [i] by Eq [ii], we obtain

N1 / N2 = e^( -5λt + λ) t = e-4λt = 1/e^4λt

But, we have given

N1 / N2 = [ 1/e ]^2 = 1/e^2

Hence, 1/e^2 = 1/e-4λt

Comparing the powers, we get

2 = 4 λt

or t = 2 / 4λt = 1 /2λ

Hence, The interval will be 4 λt !