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A radioactive element X converts into another stable element Y. Half life of X is 2 hrs. Initially only X is present. After time t, the ratio of atoms of X and Y is found to be 1 : 4, then t in hours is

Option: 1

2


Option: 2

4


Option: 3

between 4 and 6


Option: 4

None of these


Answers (1)

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Let N_{0} be the number of atoms of X at time t = 0
Then at t = 4 hrs (two half lives)

\begin{aligned} & \mathrm{N}_{\mathrm{x}}=\frac{\mathrm{N}_0}{4} \text { and } \mathrm{N}_{\mathrm{y}}=\frac{3 \mathrm{~N}_0}{4} \\ & \therefore \quad \frac{\mathrm{N}_{\mathrm{x}}}{\mathrm{N}_{\mathrm{y}}}=\frac{1}{3} \end{aligned}

and at t = 6 hrs (three half lives)

\mathrm{N}_{\mathrm{x}}=\frac{\mathrm{N}_0}{8} \text { and } \mathrm{N}_{\mathrm{y}}=\frac{7 \mathrm{~N}_0}{8} \text { or } \frac{\mathrm{N}_{\mathrm{x}}}{\mathrm{N}_{\mathrm{y}}}=\frac{1}{7}

The given ratio \frac{1}{4} lies between \frac{1}{2} and \frac{1}{7}

Therefore, t lies between 4 hrs and 6 hrs.

.

Posted by

Rishi

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