Question

A radioactive substance has 10^8 nuclei. Its half life is 30s. The no, of nuclei left after 15s is nearly

Answer 1

A radioactive for the substance has 10^8 nuclei

k=0.693/t1/2

t1/2= half life (Its half life is 30s)

k=disintegration or rot consistent  0.693/30 = k

k=0.0231

k=2.303log(a/a-x)/t

a=initial sum

a-x= sum left after deterioration

t=time of rot

0.0231=2.303log(108/108-x)/15

108/108-x=1.41

Amount left=108-x=7,09,21,986 nuclei.

Therefore The no, of nuclei left after 15s is nearly = 7,09,21,986 nuclei

Answer 2

Given:

• nuclei number $N_{0}=10^{8}$
• half life, T = 30 s

To find: number of nuclei left after 15 s = ?

Solution:

We know that, decay constant,

$\quad\quad\lambda = \frac{0.693}{T}$

$\quad\quad\quad =\frac{0.693}{30}$

$\quad\quad\quad =0.0231$

∴ the number of nuclei left after 15 s is

$\quad N=N_{0}e^{-\lambda t}$

$\to N=10^{8}\times e^{-0.0231\times 15}$

$\to \bold{N=7.07\times 10^{7}}$