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Chapter 25: Problem 21

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.

Short Answer

Expert verified
Therefore, we estimate the number of atoms of \(_{27}^{60}\) Co in the sample to be approximately \(4.4 x 10^{8}\) atoms.

Step by step solution

01

Identify Given Information

From the problem, we know that the disintegration rate (R) is 6740 disintegrations per hour, and the half-life (t1/2) of \(_{27}^{60}\) Co is 5.2 years.
02

Convert half-life to seconds

Since R is given in disintegrations per hour, it will be beneficial to convert the t1/2 from years to seconds. There are 8760 hours in a year, so \(t1/2 = 5.2 * 8760 = 45552\) hours.
03

Calculate Decay Constant

The decay constant (\( \lambda \)) is related to the half-life through the equation \( t1/2 = \frac{0.693}{\lambda} \). Solving for \( \lambda \) we get \( \lambda = \frac{0.693}{t1/2} \) which equals \(\frac{0.693}{45552} = 1.52 x 10^{-5} h^{-1} \).
04

Determine the Number of Atoms

The disintegration rate is given by R = \( \lambda*N \), where N is the number of radioactive atoms. Solving for N, we get \( N = \frac{R}{\lambda} \) which equals \( \frac{6740}{1.52 x 10^{-5}} = 4.4 x 10^{8} \) atoms.

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Most popular questions from this chapter

Each of the following isotopes is radioactive: (a) \(\frac{28}{15} \mathrm{P}\) (b) \(45 \mathrm{K} ;\) (c) \(^{73} \mathrm{Zn}\). Which would you expect to decay by 30 . \(\beta^{+}\) emission?

A sample of radioactive \(\frac{35}{16} \mathrm{S}\) disintegrates at a rate of \(1.00 \times 10^{3}\) atoms \(\min ^{-1} .\) The half-life of \(_{16}^{35} \mathrm{S}\) is \(87.9 \mathrm{d}\) How long will it take for the activity of this sample to decrease to the point of producing (a) \(253 ;\) (b) \(104 ;\) and (c) 52 dis \(\min ^{-1} ?\)

A nuclide has a decay constant of \(4.28 \times 10^{-4} \mathrm{h}^{-1}\). If the activity of a sample is \(3.14 \times 10^{5} \mathrm{s}^{-1},\) how many atoms of the nuclide are present in the sample? (a) \(2.64 \times 10^{12} ;\) (b) \(7.34 \times 10^{8}\) (c) \(2.04 \times 10^{5}\) (d) \(4.40 \times 10^{10} ;\) (e) none of these.

Complete the following nuclear equations. (a) \(\frac{23}{11} \mathrm{Na}+? \longrightarrow_{11}^{24} \mathrm{Na}+_{1}^{1} \mathrm{H}\) (b) \(_{27}^{59} \mathrm{Co}+_{0}^{1} \mathrm{n} \longrightarrow_{25}^{56} \mathrm{Mn}+?\) (c) \(?+_{1}^{2} \mathrm{H} \longrightarrow_{94}^{240} \mathrm{Pu}+_{-1}^{0} \beta\) (d) \(^{246} \mathrm{Cm}+? \longrightarrow_{102}^{254} \mathrm{No}+5_{0}^{1} \mathrm{n}\) (e) \(^{238} \mathrm{U}+? \longrightarrow_{99}^{246} \mathrm{Es}+6 \frac{1}{0} \mathrm{n}\)

Calculate the minimum kinetic energy (in megaelectronvolts) that \(\alpha\) particles must possess to produce the nuclear reaction $$_{2}^{4} \mathrm{He}+^{14}_{7} \mathrm{N} \longrightarrow^{17}_{8} \mathrm{O}+_{1}^{1} \mathrm{H}.$$ The nuclidic masses are \(_{2}^{4} \mathrm{He}=4.00260 \mathrm{u}\); \(_{7}^{14} \mathrm{He}=14.00307\mathrm{u}\);\(_{1}^{1} \mathrm{H}=1.00783 \mathrm{u}\);\(_{8}^{17} \mathrm{H}=16.99913 \mathrm{u}\);
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