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Chapter 21: Problem 80

A \(26.00-\mathrm{g}\) sample of water containing tritium, \({ }_{1}^{3} \mathrm{H},\) emits \(1.50 \times 10^{3}\) beta particles per second. Tritium is a weak beta emitter with a half-life of 12.3 yr. What fraction of all the hydrogen in the water sample is tritium?

Short Answer

Expert verified
The fraction of tritium in the water sample is approximately \(5.12 \times 10^{-21}\).

Step by step solution

01

Calculate the number of tritium atoms

To calculate the number of tritium atoms in the water sample, we will use the radioactive decay formula: \[N(t) = N_0e^{-\lambda t}\] where N(t) is the number of remaining radioactive atoms at time t, N0 is the initial number of radioactive atoms, λ (lambda) is the decay constant, and t is the time in years. We know the half-life of tritium (T) is 12.3 years. We can find the decay constant, λ, using the following equation: \[\lambda = \frac{\ln(2)}{T}\] Then, we can find the initial number of tritium atoms (N0) using the given beta particles emitted per second (R) using the equation: \[R = \lambda N_0\]
02

Calculate the decay constant of tritium

Using the half-life of tritium (T = 12.3 yr), we can find the decay constant λ: \[\lambda = \frac{\ln(2)}{12.3} \approx 0.0563 \,\text{yr}^{-1}\]
03

Find the initial number of tritium atoms

The beta particles emitted per second (R) is given as 1.50 x 10^3. Using the decay constant we calculated in step 2, we can find the initial number of tritium atoms (N0): \[N_0 = \frac{R}{\lambda} = \frac{1.50 \times 10^{3}}{0.0563} \approx 2.668 \times 10^{4}\] So, there are approximately 2.668 x 10^4 tritium atoms in the water sample.
04

Calculate the mass of tritium in the water sample

The molar mass of tritium is 3 g/mol. Using Avogadro's number (6.022 x 10^23 atoms/mol), we can find the mass of tritium in the water sample: Mass of tritium = (2.668 x 10^4 atoms) x (3 g/mol) / (6.022 x 10^23 atoms/mol) ≈ 1.33 x 10^(-19) g
05

Calculate the fraction of tritium in the water sample

Given that the total mass of the water sample is 26.00 g, we can now find the fraction of tritium in the water sample by dividing the mass of tritium by the total mass of the water sample: Fraction of tritium = (1.33 x 10^(-19) g) / (26.00 g) ≈ 5.12 x 10^(-21) The fraction of tritium in the water sample is approximately 5.12 x 10^(-21).

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Most popular questions from this chapter

Write balanced equations for (a) \({ }_{92}^{238} \mathrm{U}(\alpha, \mathrm{n}){ }_{94}^{241} \mathrm{Pu},\) (b) \({ }_{7}^{14} \mathrm{~N}(\alpha, \mathrm{p}){ }^{17} \mathrm{O}\) (c) \({ }_{26}^{56} \mathrm{Fe}(\alpha, \beta)_{29}^{60} \mathrm{Cu}\).

One nuclide in each of these pairs is radioactive. Predict which is radioactive and which is stable: (a) \({ }_{20}^{40} \mathrm{Ca}\) and \({ }_{20}^{45} \mathrm{Ca},(\mathbf{b}){ }^{12} \mathrm{C}\) and \({ }^{14} \mathrm{C},(\mathrm{c})\) lead- 206 and thorium- \(230 .\) Explain your choice in each case.

Which are classified as ionizing radiation: X-rays, alpha particles, microwaves from a cell phone, and gamma rays?

Decay of which nucleus will lead to the following products: (a) bismuth-211 by beta decay; (b) chromium-50 by positron emission; (c) tantalum-179 by electron capture; (d) radium226 by alpha decay?

Using the concept of magic numbers, explain why alpha emission is relatively common, but proton emission is nonexistent.
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