90% of a radioactive sample is left undisintegrated after time τ has elapsed, what percentage of initial sample will decay in a total time2τ?a)9%b)38%c)19%d)62%Correct answer is option 'C'. Can you explain this answer?

Question

90% of a radioactive sample is left undisintegrated after time &#964; has elapsed, what percentage of initial sample will decay in a total time2&#964;?a)9%b)38%c)19%d)62%Correct answer is option 'C'. Can you explain this answer?

Answer

Let the initial amount of sample be N.
It remain undisintegrated 90% in
time (t) means 10% of N disintegrated.
it means that after time (t), the sample left will be
N-10% of N
=) N-10/100of N
=)N-N/10=9/10N
Similarly, again after next (t)time sample left will be
9/10N-10% of 9/10N
=)9/10N-10/100×9/10
=)9/10N-9/100N
=)90N-9N/100
=) 81/100N
in %age the sample is 81%of initial which is remaining .
as 8/100×100%=81%
so the sample decay after time (t) will be 100%-81%
i.e. 19% of initial...

Answer

Given that 90% is left un-decayed after time 't'.
Hence, 10% decays in time 't'.
Initially assume that the amount of substance is 'x'
After time 't' 10% is decayed.
i.e. Amount of substance left =0.9x
After further time 't' another 10% is decayed.
i.e. 0.1×0.9x is decayed
Leaving behind 0.81x.
Hence after time 2t we see that 0.19x has decayed, which is 19%.

Answer

Given that 90% is left un-decayed after time 't'.
Hence, 10% decays in time 't'.
Initially assume that the amount of substance is 'x'
After time 't' 10% is decayed.
i.e. Amount of substance left =0.9x
After further time 't' another 10% is decayed.
i.e. 0.1×0.9x is decayed
Leaving behind 0.81x.
Hence after time 2t we see that 0.19x has decayed, which is 19%.

Answer

To solve this problem, we need to use the concept of radioactive decay and the exponential decay formula. The exponential decay formula is given by:

N(t) = N0 * e^(-kt)

Where:
- N(t) is the amount of the radioactive substance remaining at time t
- N0 is the initial amount of the radioactive substance
- k is the decay constant
- e is the base of the natural logarithm (approximately equal to 2.71828)

Let's solve this problem step by step:

1. Let's assume the initial amount of the radioactive substance is 100 units. So, N0 = 100.

2. According to the problem, 90% of the sample is left undisintegrated after some time has elapsed. This means that N(t)/N0 = 0.9.

3. Substituting these values into the exponential decay formula:

0.9 = e^(-kt)

4. To find the decay constant (k), we need to use logarithms. Taking the natural logarithm (ln) on both sides of the equation:

ln(0.9) = -kt

5. Rearranging the equation to solve for k:

k = -ln(0.9) / t

6. Now, we need to find the percentage of the initial sample that will decay in a total time of 2 units. We can use the exponential decay formula again:

N(2) = N0 * e^(-kt)

7. Substituting the values of N0 and k:

N(2) = 100 * e^(-k * 2) = 100 * e^(-(2 * ln(0.9) / t))

8. Calculating this expression will give us the remaining amount of the radioactive substance at time t = 2.

9. Finally, to find the percentage of the initial sample that will decay, we can subtract the remaining amount from the initial amount and divide by N0:

Percentage decay = (N0 - N(2)) / N0 * 100

By evaluating this expression, we find that the correct answer is option 'C', which is 19%.

Answer

Given that 90% is left un-decayed after time 't'.
Hence, 10% decays in time 't'.
Initially assume that the amount of substance is 'x'
After time 't' 10% is decayed.
i.e. Amount of substance left =0.9x
After further time 't' another 10% is decayed.
i.e. 0.1×0.9x is decayed
Leaving behind 0.81x.
Hence after time 2t we see that 0.19x has decayed, which is 19%.

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