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Chapter 21: Problem 53

Tritium, \({ }^{3} \mathrm{H},\) is radioactive and decays by electron emission. Its half-life is 12.5 yr. In ordinary water the ratio of \({ }^{1} \mathrm{H}\) to \({ }^{3} \mathrm{H}\) atoms is \(1.0 \times 10^{17}\) to \(1 .\) (a) Write a balanced nuclear equation for tritium decay. (b) How many disintegrations will be observed per minute in a \(1.00-\mathrm{kg}\) sample of water?

Short Answer

Expert verified
The decay equation for Tritium is \({ }^{3} \mathrm{H} \rightarrow { }^{3} \mathrm{He} + -1^0 e \), and there would be roughly 1.06 * 10^{3} disintegrations per minute in a 1.00 kg sample of water.

Step by step solution

01

Writing the decay equation for Tritium

Tritium, which is a hydrogen isotope with two neutrons (\({ }^{3} \mathrm{H}\)), decays through beta decay into Helium-3 (\({ }^{3} \mathrm{He}\)) by emitting an electron (\(-1^0 e\)). The balanced decay equation would therefore be: \({ }^{3} \mathrm{H} \rightarrow { }^{3} \mathrm{He} + -1^0 e \)
02

Calculating the number of Hydrogen atoms in water

Since water (H2O) has two Hydrogen atoms, in 1 kg of water, which is roughly 55.5 moles (since the molar mass of water is approximately 18 g/mol), we would have \((2 * 6.022 * 10^{23} * 55.5) = 6.68 * 10^{25}\) Hydrogen atoms.
03

Finding the number of Tritium atoms

Given the ratio of Hydrogen to Tritium atoms is \(1.0 \times 10^{17}\) to \(1\), the number of Tritium atoms would be \((6.68 * 10^{25}) / (1.0 \times 10^{17}) = 6.68 * 10^{8}\) atoms.
04

Calculating the rate of disintegrations

Tritium's decay can be described by the expression N=N0*0.5^(t/T), where N0 is initial number of nuclei, t is time, and T is half-life. The rate of disintegration (activity) is given by -dN/dt = N0*ln(2)/T* 0.5^(t/T) For the activity per minute (when t=0), the equation simplifies to -dN/dt = N0*ln(2)/T = (6.68 * 10^{8} atoms)*(0.693)/(12.5*365.25*24*60) = 1.06 * 10^{3} disintegrations per minute.
05

Rounding and writing the final answer

Concluding from above calculation, we should round the result to the appropriate number of significant figures. Seeing the data we have, we should round the final result to three significant digits, so the solution is 1.06 * 10^{3} disintegrations per minute.

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Most popular questions from this chapter

What are the steps in balancing nuclear equations?

(a) Calculate the energy released when a \({ }^{238} \mathrm{U}\) isotope decays to \({ }^{234} \mathrm{Th} .\) The atomic masses are given by: \(^{238} \mathrm{U}: 238.0508 \mathrm{amu} ;{ }^{234} \mathrm{Th}: 234.0436 \mathrm{amu} ;{ }^{4} \mathrm{He}\) 4.0026 amu. (b) The energy released in (a) is transformed into the kinetic energy of the recoiling \({ }^{234} \mathrm{Th}\) nucleus and the \(\alpha\) particle. Which of the two will move away faster? Explain.

Why is it preferable to use nuclear binding energy per nucleon for a comparison of the stabilities of different nuclei?

Calculate the nuclear binding energy (in J) and the nuclear binding energy per nucleon of the following isotopes: (a) \({ }_{3}^{7} \mathrm{Li}(7.01600 \mathrm{amu})\) and (b) 35 17 Cl \((34.95952 \mathrm{amu})\)

(a) What is the activity, in millicuries, of a 0.500 -g sample of \({ }_{93}^{237} \mathrm{~Np}\) ? (This isotope decays by \(\alpha\) -particle emission and has a half-life of \(2.20 \times 10^{6}\) yr.) (b) Write a balanced nuclear equation for the decay of \({ }^{237} \mathrm{~Np}\)
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