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Fundamentals Of Physics
Found in: Page 1304

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Short Answer

Figure 42-19 shows the decay of parents in a radioactive sample. The axes are scaled by Ns=2.00x106 and tS=10.0s. What is the activity of the sample at?

The activity of the sample at t = 27s is 60 Bq.

See the step by step solution

Step by Step Solution

Step 2: Determine the formula for the decay

Write the formula for undecayed sample remaining after a given time, as follows:

N=N0e-λt …… (i)

Write the formula for the rate of decay:

R=λN

Here, λ is the disintegration constant, N is the number of undecayed nuclei.

Step 3: Calculate the activity of the sample at 27h

The number of atoms present initially at t = 0 is N0=2×106. From Fig. 42-19, determine that the number is halved at t = 2s. Thus, using equation (i), determine the disintegration constant of the sample as follows:

λ=1tlnN0N =12slnN0N02 =0.3466 s-1

Now, using this value in equation (i), determine value of the remaining nuclei of the radioactive sample at t=27s as follows:

N=2×106e-0.3466s-127 s =173

Now, the activity of the sample at the given time is given using the above values in equation (ii) as follows:

R=0.3466s-1173 60 s-1 =60 Bq

Hence, the activity of the sample is 60 Bq.

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