Question

the half-life of a radioactive substance is 30 days. what time will it take for(3/4)th of its original mass to disintegrate?

Answer 1

After 12.45 days radioactive substance will (3/4)th of its original mass.

Explanation:

The half-life of a radioactive substance is 30 days.

Formula for half life:-

$N(t)=N_o(0.5)^{t/t_{1/2}}$

where,

• $N_0$ initial radioactive substance.

• N(t) is final substance.

• $t_{1/2}$ half life of the substance.

$N(t)=\dfrac{3}{4}N_0$

$\dfrac{3}{4}N_0=N_o(0.5)^{t/30}$

Apply ln both sides

$\ln(0.75)\dfrac{t}{30}\ln (0.5)$

$t=\dfrac{30\ln(0.75)}{\ln(0.5)}$

$t=12.45\text{ days}$

hence, After 12.45 days radioactive substance will (3/4)th of its original mass.

#Learn more:

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Answer 2

It would require 60 days to disintegrate 3/4th of the original mass of the radioelement.

Explanation:

Given, half-life of a radioactive substance = 30 days.

From formula for half life, we have  

N(t) = N(o)(0.5)^(t/root of t)

Please refer picture for the correct formatting of the formula.

N(o) is the original quantity

N(t) is quantity that still remains = 1/4 N(o)

root of t is the half-life of the substance = 30 days

Applying the values, we get:

1/4 N(o) = N(o)(0.5)^(t/30)

Applying In, we get  

In(0.25)*(t/30)*In(0.5)

t = 30 * In(0.25) / In(0.5)

 = 60 days