
After five half-lives, percentage of original radioactive atoms left is:
A. $1\% $
B. $0.3\% $
C. $3.125\% $
D. $0.2\% $
Answer
576.9k+ views
Hint: Half-life in the time taken by any quantity to reduce to half of its original size.
Complete step by step answer:
We know that, after a half-life, only half of the quantity of any particle is left.
Let us assume, the initial concentration of the radioactive atom is$100\% $.
After the first half of it
i.e.$\dfrac{1}{2} \times 100\% = 50\% $
After the second half life, it will reduce to half of remaining concentration.
i.e.$\dfrac{1}{2} \times 50\% = 25\% $
After the third half-life, it will reduce to half of the remaining concentration.
i.e.$\dfrac{1}{2} \times 25\% = 12.5\% $
After the fourth half-life, it will reduce to half and the remaining concentration.
i.e.$\dfrac{1}{2} \times 12.5\% = 6.25\% $
After the fifth half-life, it will reduce to half of the remaining concentration.
$\dfrac{1}{2} \times 6.25\% = 3.125\% $
Therefore, from the above explanation the correct option is (C)$3.125\% $.
Note: From this question, we can observe that, in every half-life, the concentration of a particle is reducing to half of its last concentration. That means, it we wait for it to reduce to exactly $0\% $, then it would take infinite half livers as there would always be some part left to the half of.
Complete step by step answer:
We know that, after a half-life, only half of the quantity of any particle is left.
Let us assume, the initial concentration of the radioactive atom is$100\% $.
After the first half of it
i.e.$\dfrac{1}{2} \times 100\% = 50\% $
After the second half life, it will reduce to half of remaining concentration.
i.e.$\dfrac{1}{2} \times 50\% = 25\% $
After the third half-life, it will reduce to half of the remaining concentration.
i.e.$\dfrac{1}{2} \times 25\% = 12.5\% $
After the fourth half-life, it will reduce to half and the remaining concentration.
i.e.$\dfrac{1}{2} \times 12.5\% = 6.25\% $
After the fifth half-life, it will reduce to half of the remaining concentration.
$\dfrac{1}{2} \times 6.25\% = 3.125\% $
Therefore, from the above explanation the correct option is (C)$3.125\% $.
Note: From this question, we can observe that, in every half-life, the concentration of a particle is reducing to half of its last concentration. That means, it we wait for it to reduce to exactly $0\% $, then it would take infinite half livers as there would always be some part left to the half of.
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