Question

Half life of a radioactive decay of c is 5730years.How much time it will take so that 25% of c was found in sample

Answer 1

Answer:

it mean 75% of sub has finished mean 2hslf life req so 5730 ×2 = 11460

Answer 2

Explanation:

It is given:

Charcoal has the initial activity, which is denoted as A_{0}=15.3A

0

=15.3 disintegrations per minute per gram.

Charcoal has the half-life, T 12=5730T12=5730 years

After a few years, the charcoal’s final activity, A = 12.3 disintegrations per minute per gram

Constant of disintegration,

\lambda=0.693 \mathrm{T} 12=0.6935370 \mathrm{y}-1λ=0.693T12=0.6935370y−1

For the action to attain 12.3 disintegrations per minute per gram, let the time taken by the sample at a time of t year.

Sample’s activity,

A=A O e-\lambda tA=AOe−λt

A=A O e-0.6935730 \times tA=AOe−0.6935730×t

\Rightarrow \mathrm{t}=1804.3 \text { years }⇒t=1804.3 years