Question

. At t=0, a radioactive substance has a mass m. Its half-life is 10 minutes. When t= t1,
the amount of substance disintegrated is m/5 and when t= t2, the amount of substance
disintegrated is 3m/5. Then the time interval (t2 - t1)is

(1) 10 minutes
(2) 20 minutes
(3) 5 minutes
(4) 7 minutes​

Answer 1

Answer:

7 I hope it is useful for you

Answer 2

Given: At t=0, a radioactive substance has a mass m. Its half-life is 10 minutes. When t= $t_1$, the amount of substance disintegrated is $\dfrac{m}{5}$ and when t=$t_2$ , the amount of substance

disintegrated is $\dfrac{3m}{5}$.

To find: We have to find the time interval $(t_2-t_1)$

Solution:

The half life of a radioactive substance can be given as-

$= \frac{0.693}{k}$

Where k is the rate constant.

The value of half life period is 10 minute so, k will be-

$\frac{0.693}{10 } \\ = 0.0693$

When t= $t_1$, the amount of substance disintegrated is $\dfrac{m}{5}$

The substance left will be-

$m - \frac{m}{5} \\ = \frac{4m}{5}$

$t_1 = \frac{1}{0.0693} ln( \frac{m}{ \frac{4m}{5} } ) \\ = \frac{1}{0.0693} ln( \frac{5}{ 4} ) \\ = 3$

and when t=$t_2$ , the amount of substance disintegrated is $\dfrac{3m}{5}$.

The amount of substance left will be-

$m - \frac{3m}{5} \\ = \frac{2m}{5}$

Now the value of $t_2$ will be-

$t_2 = \frac{1}{0.0693} ln( \frac{m}{ \frac{2m }{5} } ) \\ = \frac{1}{0.0693} ln( \frac{5}{2} ) \\ = 13$

$(t_2-t_1) = 13 - 3 \\ = 10$

The time interval is 10 minutes.