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Chapter 25: Problem 89

A nuclide has a decay rate of \(2.00 \times 10^{10} \mathrm{s}^{-1} .\) After 25.0 days, its decay rate is \(6.25 \times 10^{8} \mathrm{s}^{-1}\). What is the nuclide's half-life? (a) 25.0 d; (b) 12.5 d; (c) 50.0 d; (d) \(5.00 \mathrm{d} ;\) (e) none of these.

Short Answer

Expert verified
The half-life of the nuclide is approximately 6.63 days. The correct answer is (e) none of these.

Step by step solution

01

Identify the Knowns

The initial decay rate \(N_0\) is \(2.00 \times 10^{10} \mathrm{s}^{-1}\). The decay rate after 25 days, denoted \(N\), is \(6.25 \times 10^{8} \mathrm{s}^{-1}\). The time \(t\) is 25 days, but in seconds is \(25 \times 24 \times 60 \times 60 = 2160000 \mathrm{s}\). This converts the time into seconds to match the decay rate units.
02

Use the Exponential Decay Formula

The exponential decay formula is \( N = N_0 \times \exp(-\lambda t)\), where \(N_0\) and \(N\) are initial and final decay rates, \(t\) is time, and \(\lambda\) is the decay constant. We need to solve for \(\lambda\) using the given values: \(6.25 \times 10^{8} = 2.00 \times 10^{10} \times \exp(-\lambda \times 2160000)\).
03

Solve for the Decay Constant

Rearranging the formula and taking the natural logarithm of both sides gives: \(\lambda = - \frac{\ln(6.25 \times 10^{8} / 2.00 \times 10^{10})}{2160000}\). When calculated, this gives \(\lambda = 1.21 \times 10^{-6} \mathrm{s}^{-1}\).
04

Derive Half-Life

The half-life \(T_{1/2}\) is related to the decay constant by \(T_{1/2} = \frac{0.693}{\lambda}\). Substituting the value of \(\lambda\) gives us \(T_{1/2} = \frac{0.693}{1.21 \times 10^{-6}}\). When calculated, this gives \(T_{1/2} = 572600 \mathrm{s}\).
05

Convert Half-Life into Days

Finally, convert the half-life into days: \(T_{1/2} = \frac{572600}{24 \times 60 \times 60}\). When calculated, this gives \(T_{1/2} = 6.63\) days. So the correct answer is not among the options given.

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Most popular questions from this chapter

232 Ra has a half-life of 11.4 d. How long would it take for the radioactivity associated with a sample of \(^{232} \mathrm{Ra}\) to decrease to \(1 \%\) of its current value?

The disintegration rate for a sample containing \(_{27}^{60} \mathrm{Co}\) as the only radioactive nuclide is 6740 dis \(\mathrm{h}^{-1}\). The half-life of 20 Co is 5.2 years. Estimate the number of atoms of \(_{27}^{60}\) Co in the sample.

Explain why the rem is more satisfactory than the rad as a unit for measuring radiation dosage.

Hydrogen gas is spiked with tritium to the extent of \(5.00 \%\) by mass. What is the activity in curies of a \(4.65 \mathrm{L}\) sample of this gas at \(25.0^{\circ} \mathrm{C}\) and 1.05 atm pressure? [Hint: Use 3.02 u as the atomic mass of tritium and data from elsewhere in the text, as necessary.]

In some cases, the most abundant isotope of an element can be established by rounding off the atomic mass to the nearest whole number, as in \(^{39} \mathrm{K},^{85} \mathrm{Rb}\), and \(^{88} \mathrm{Sr}\). But in other cases, the isotope corresponding to the rounded-off atomic mass does not even occur naturally, as in \(^{64} \mathrm{Cu}\). Explain the basis of this observation.
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