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Standard XII

Physics

Nuclear Fission

A radioactive...

Question

A radioactive substance is decaying with $t_{1 / 2} = 30$ days. On being separated into two fractions, one of the fractions, immediately after separation, decays with $t_{1 / 2} = 2$ days. The other fraction, immediately after separation would show:

constant activity

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increasing activity

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decay with

t_{1 / 2} = 30

days

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decay with

t_{1 / 2} = 28

days

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Solution

The correct option is B increasing activity
From basic radioactivity, we know that separation of nuclei, giving one finite $t_{1 / 2}$ would lead to another unstable (in this case) nuclei.
So option B correct.

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Similar questions

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The half life of a radio active substance is $20$ days. If $\frac{2}{3}$ part of the substance has decayed in time $t_{2}$ and $\frac{1}{3}$ part of it has decayed in time $t_{1}$ then the time interval between $t_{2}$ and $t_{1} i s (t_{2} - t_{1}) =$

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A radioactive substance is decaying with t1/2=30 days. On being separated into two fractions, one of the fractions, immediately after separation, decays with t1/2=2 days. The other fraction, immediately after separation would show:

The correct option is B increasing activityFrom basic radioactivity, we know that separation of nuclei, giving one finite t1/2 would lead to another unstable (in this case) nuclei.So option B correct.

A radioactive substance is decaying with $t_{1 / 2} = 30$ days. On being separated into two fractions, one of the fractions, immediately after separation, decays with $t_{1 / 2} = 2$ days. The other fraction, immediately after separation would show:

The correct option is B increasing activity
From basic radioactivity, we know that separation of nuclei, giving one finite $t_{1 / 2}$ would lead to another unstable (in this case) nuclei.
So option B correct.