Question 140(ii)

Consider a quantity of a radioactive substance. The fraction of this quantity that remains after t half-lives can be found by using the expression $3^{- t}$ .
After how many half-lives will the fraction be $\frac{1}{243}$ of the original?

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Solution

Given, the fraction of a radioactive substance that remains after t half-lives can be found by using the expression $3^{- t}$ .

So,
$\frac{1}{243} = 3^{- t} \Rightarrow \frac{1}{3^{5}} = \frac{1}{3^{t}} = 3^{- 5} [∵ 3 \times 3 \times 3 \times 3 \times 3 = 3^{5} a n d a^{- m} = \frac{1}{a^{m}}]$

On comparing both sides, we get t = 5 half-lives.

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Question 140(ii)

Consider a quantity of a radioactive substance. The fraction of this quantity that remains after t half-lives can be found by using the expression $3^{- t}$ .
After how many half-lives will the fraction be $\frac{1}{243}$ of the original?

Question 140(i)

Consider a quantity of a radioactive substance. The fraction of this quantity that remains after t half-lives can be found by using the expression $3^{- t}$
What frcation of the substance remains after 7 half-lives?

Question 140(i)

Question 139

A half-life is the amount of time that it takes for a radioactive substance to decay one-half of its original quantity.
Suppose radioactive decay causes 300 grams of a substance to decrease $300 \times 2^{- 3}$ grams after 3 half-lives. Evaluate $300 \times 2^{- 3}$ to determine how many grams of the substance is left.
Explain why the expression $300 \times 2^{- n}$ can be used to find the amount of the substance that remains after n half-lives.

Q. The percentage of quantity of a radioactive material that remains after

5

half-lives will be

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Question 140(ii) Consider a quantity of a radioactive substance. The fraction of this quantity that remains after t half-lives can be found by using the expression 3−t. After how many half-lives will the fraction be 1243 of the original?

Given, the fraction of a radioactive substance that remains after t half-lives can be found by using the expression 3−t. So, 1243=3−t⇒135=13t=3−5 [∵3×3×3×3×3=35 and a−m=1am] On comparing both sides, we get t = 5 half-lives.

Question 140(ii)

Consider a quantity of a radioactive substance. The fraction of this quantity that remains after t half-lives can be found by using the expression $3^{- t}$ .
After how many half-lives will the fraction be $\frac{1}{243}$ of the original?