Question

The number of nuclei of a radioactive sample becomes 1/8th part of its initial value N, in period of 6 years. The mean life (in years) of the radioactive sample will be
1) 6/ln2
2) 4/ln2
3) 3/2ln2
4) 2/ln2​

Answer 1

Explanation:

activity of radioactive substance at any time t is given by

⇒N=N

0

e

−λt

given that

N

0

N

=

16

1

at time t=40days

⇒

16

1

=e

−40λ

⇒e

40λ

=16

⇒λ=

40

ln16

we know that

t

2

1

=

λ

ln2

after putting the value of λ in above equation

⇒t

2

1

=

ln16

ln2

×40=10

so the correct option will be (B).

Answer 2

Answer:

Correct anser will be option 4) 2/ln2 years

Explanation:

Step 1 :

Mean life of the radioactive sample will be,

$\tau = \frac{1}{\lambda}$

Step 2 :

From decaying of a radioactive sample we know that,

$N = N_0 e^{-\lambda t}$

Step 3 :

Give that number of nuclei becomes 1/8 th of its initial value

$N = \frac{N_0}{8}$

Time t = 6 years

Putting in the second equation -

$N = N_0 e^{-\lambda t}$

$\frac{N_0}{8} = N_0 e^{-6\lambda}$

$\frac{1}{8} = e^{-6\lambda}$

$e^{6\lambda} = 8$

$6\lambda = ln8\\6\lambda = 3ln2\\\lambda = \frac{ln2}{2}$

Step 4 :

Putting value of λ in first equation we get mean life in years,

$\tau = \frac{2}{ln2}$   years