Question

A radioactive substance has a half-life of 10 hrs. What percentage of it will be left after 40

hrs? ​

Answer 1

Given : A radioactive substance has a half life of 10 hrs.

To find : the percentage of it will be after 40 hours

solution : half life of radioactive decay is given by, $T_{1/2}=\frac{ln2}{\lambda}$

where, $\lambda$ decay constant.

here, $T_{1/2}$ = 10 hrs

so, 10 = $\frac{ln2}{\lambda}$

⇒$\lambda=\frac{ln2}{10}$ hr¯¹ ......(1)

Now using formula of radioactive decay at time t,

$N(t)=N_0e^{-\lambda t}$

here $N_0$ is the initial quantity of radioactive substance and N(t) is the remaining quantity after time t.

here t = 40 hrs, $\lambda=\frac{ln2}{10}$

so, $N(t)=N_0e^{-\frac{ln2}{10}\times40}$

$=N_0e^{-4ln2}$

$=N_0e^{ln2^{-4}}$

$=\frac{N_0}{2^4}$

⇒$\frac{N(t)}{N_0}=\frac{1}{2^4}$

percentage of it will left after 40 hrs = $\frac{N(t)}{N_0}\times100=\frac{100}{16}$

= 6.25 %

Therefore the percentage of radioactive substance will left after 40 hrs is 6.25 %.

shortcut : half life is 10hrs . means if 100 is amount of substance then 50 will left after 10 hrs,

25 will left after 20 hrs

12.5 will left after 30 hrs

6.25 will left after 40 hrs .

hence, 6.25 % will left after 40 hrs.

Answer 2

Percentage of the substance left after 40 hours = 6.25%

Explanation:

Given

Half life of the radioactive substance

$t_{1/2}=10$ hrs

Period $T=40$ hrs

We know that

Radioactive substance left after T period

$N=N_0(\frac{1}{2})^{T/t_{1/2}}$           (where N₀ is initial amount)

$\implies N=N_0(\frac{1}{2})^{40/10}$

$\implies \frac{N}{N_0}=\frac{1}{16}$

Percentage Left

$=\frac{N}{N_0}\times 100$

$=\frac{1}{16}\times 100$

$=6.25\%$

Hope this answer is helpful.

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