wiz-icon
MyQuestionIcon
MyQuestionIcon
2
You visited us 2 times! Enjoying our articles? Unlock Full Access!
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?


Open in App
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.


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Half and Average Life
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon