A sample of radioactive material contains 1018atoms. The half life of the material is 2 days, then the activity of the sample isa)3.5 x 1014Bqb)3.5 x 1012Bqc)7 x 1011Bqd)7 x 1016BqCorrect answer is option 'B'. Can you explain this answer?

Question

A sample of radioactive material contains 1018atoms. The half life of the material is 2 days, then the activity of the sample is&#8203;a)3.5 x 1014Bqb)3.5 x 1012Bqc)7 x 1011Bqd)7 x 1016BqCorrect answer is option 'B'. Can you explain this answer?

Answer

To find activity of the sample --->which is the rate of disintegration.

Since radioactivity comes under 1^o kinetics.

[R]=k[A] [A]-->amount of initial sample 10¹⁸ atoms

Given,

Half-life=2days

K=0.693/2x24x60x60 sec

R=(0.693/2x24x60x60)x10¹⁸

R≈3.5x10¹² Bq

Answer

Answer

Calculation of Activity of Radioactive Material

Given:

Number of atoms in the sample N = 1018

Half-life of the material t1/2 = 2 days

We know that the radioactive decay law is given by:

N = N0 x (1/2)^(t/t1/2)

where,
N0 is the initial number of atoms
t is the time elapsed since the initial number of atoms

Activity of a radioactive sample is defined as the rate of decay of the sample, which is given by:

A = λN

where,
λ is the decay constant

We know that the decay constant is related to the half-life by the equation:

λ = ln(2)/t1/2

Substituting the given values, we get:

λ = ln(2)/2 = 0.3466 day^-1

Now, substituting the values of N and λ in the equation for activity, we get:

A = λN = 0.3466 x 1018 = 3.5 x 10^12 Bq

Therefore, the activity of the radioactive material is 3.5 x 10^12 Bq (option B).

Answer

C

Answer

Answer

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