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Half Life and Average Life

At any instan...

Question

At any instant, the ratio of the amount of radioactive substances is $2 : 1$ . If their half life be respectively $12$ and $16$ hours, then after two days, what will be the ratio of the substances?

1 : 1

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2 : 1

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1 : 2

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1 : 4

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Solution

The correct option is B $1 : 1$
For substance A :
$2 N_{0} \to N_{A} = 2 N_{0} {(\frac{1}{2})}^{48 / 12} = \frac{N_{0}}{2^{3}} = \frac{N_{0}}{8}$
For substance B:
$N_{0} \to N_{B} = 2 N_{0} {(\frac{1}{2})}^{48 / 16} = \frac{N_{0}}{2^{3}} = \frac{N_{0}}{8}$
$N_{A} : N_{B} = 1 : 1$

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At any instant, the ratio of the amount of radioactive substances is 2:1. If their half life be respectively 12 and 16 hours, then after two days, what will be the ratio of the substances?

The correct option is B 1:1For substance A :2N0→NA=2N0(12)48/12=N023=N08For substance B:N0→NB=2N0(12)48/16=N023=N08NA:NB=1:1

At any instant, the ratio of the amount of radioactive substances is $2 : 1$ . If their half life be respectively $12$ and $16$ hours, then after two days, what will be the ratio of the substances?

The correct option is B $1 : 1$
For substance A :
$2 N_{0} \to N_{A} = 2 N_{0} {(\frac{1}{2})}^{48 / 12} = \frac{N_{0}}{2^{3}} = \frac{N_{0}}{8}$
For substance B:
$N_{0} \to N_{B} = 2 N_{0} {(\frac{1}{2})}^{48 / 16} = \frac{N_{0}}{2^{3}} = \frac{N_{0}}{8}$
$N_{A} : N_{B} = 1 : 1$