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One microgram of radioactie sodium Na^(2...

One microgram of radioactie sodium `Na^(24)` with a half life of 15 h was injected fin to a living system for a bio assay .How long will iot take for the radioactivity to fall to 25% of the initial value?

A

60 h

B

22.5 h

C

375 h

D

30h

Text Solution

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The correct Answer is:
To solve the problem of how long it will take for the radioactivity of sodium-24 (Na-24) to fall to 25% of its initial value, we can follow these steps: ### Step 1: Understand the Problem We know that the half-life of Na-24 is 15 hours. We need to find the time it takes for the radioactivity to reduce to 25% of its initial value. ### Step 2: Identify Initial and Final Values - Initial amount of Na-24 = 1 microgram - Final amount after decay = 25% of initial = 0.25 microgram ### Step 3: Use the Concept of Half-Life The half-life is the time required for a quantity to reduce to half its initial value. After one half-life (15 hours), the amount of Na-24 will be: - After 15 hours: 0.5 microgram (50% of initial) - After another half-life (30 hours total): 0.25 microgram (25% of initial) ### Step 4: Calculate the Total Time Since we need to go from 1 microgram to 0.25 microgram, we can see that this requires two half-lives: - First half-life (15 hours): 1 microgram → 0.5 microgram - Second half-life (15 hours): 0.5 microgram → 0.25 microgram Thus, the total time required is: \[ \text{Total time} = 15 \text{ hours} + 15 \text{ hours} = 30 \text{ hours} \] ### Conclusion It will take **30 hours** for the radioactivity of Na-24 to fall to 25% of its initial value. ---

To solve the problem of how long it will take for the radioactivity of sodium-24 (Na-24) to fall to 25% of its initial value, we can follow these steps: ### Step 1: Understand the Problem We know that the half-life of Na-24 is 15 hours. We need to find the time it takes for the radioactivity to reduce to 25% of its initial value. ### Step 2: Identify Initial and Final Values - Initial amount of Na-24 = 1 microgram - Final amount after decay = 25% of initial = 0.25 microgram ...
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