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Question

After 280 days, the activity of a radioactive sample is =6000dps. The activity reduces to3000dps after another 140 days. What Is The initial activity of the sample in dps?


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Solution

Step 1. Given data

Activity reduces from 6000 to 3000dps i. e. half.

T12=140 days

280 days =2 half life's

After 280 days =6000dps activity

Step 2. Concept to be used

The decays constant λ is the inverse of the mean lifetime (average lifetime of a radioactive particle before decay).

So, the rate equation in first order reaction is,

λ=2.303tlogA0A

Here, A0 is the initial activity and A is the activity at time t.

Step 3. Calculate the initial activity

Now, we will taking the number of days as time.

Substitute the values in the above expression, we get,

λ=2.303280logA06000 -------- 1

and

λ=2.303420logA03000 ------- 2

Equate the both equations 1 and 2, we get,

2.303280logA06000=2.303420logA03000

logA06000=11.5logA03000

Step 4. Solve by using exponential rule.

Take the exponential on both sides and solving the above equation we get,

logA06000=11.5logA03000

Multiply both sides by 1.5.

So,

log10A06000×1.5=11.5log10A03000×1.5

log10A06000×1.5=log10A03000

Use log rules logaxn=nlogax, we get,

0.000166666A01.5=A03000

multiply both sides by 3000.

So,

0.000166666A01.5×3000=A03000×3000

0.000166666A01.5×3000=A0

Take both sides of the equation to the power of 2, we get,

0.000166666A01.5×30002=A02

0.000041666A03=A02

Substract A02 from both sides, we get,

0.000041666A03-A02=A02-A02

0.000041666A03-A02=A02-A02

0.000041666A0-1=0

Step 5. Use zero factor principle

If ab=0 then a=0 or b=0.

So,

0.000041666A0-1=0

A0=10.000041666

A0=24000dps

Hence, the initial activity is 24,000dps.


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