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Chapter 21: Problem 36

It takes 180 minutes for a 200 -mg sample of an unknown radioactive substance to decay to \(112 \mathrm{mg}\). What is the halflife of this substance?

Short Answer

Expert verified
The half-life of the unknown radioactive substance is approximately \(324.84\) minutes.

Step by step solution

01

Write down the decay formula to be used

The radioactive decay formula is given by: \[ N(t) = N_{0} \left(\frac{1}{2}\right)^\frac{t}{t_{1/2}} \] Where \(N(t)\) is the remaining amount of the substance after time \(t\), \(N_{0}\) is the initial amount of the substance, \(t_{1/2}\) is the half-life, and \(t\) is the time of decay.
02

Plug in the given information into the decay formula

We are given that the initial mass (\(N_{0}\)) is 200 mg, the final mass (\(N(t)\)) is 112 mg, and the decay time (\(t\)) is 180 minutes. We need to find the half-life (\(t_{1/2}\)). Plug these values into the decay formula: \[ 112 = 200 \left(\frac{1}{2}\right)^\frac{180}{t_{1/2}} \]
03

Solve for t_{1/2}

To find \(t_{1/2}\), we will first divide both sides of the equation by 200: \[ \frac{112}{200} = \left(\frac{1}{2}\right)^\frac{180}{t_{1/2}} \] Next, take the logarithm base \( \frac{1}{2} \) of both sides: \[ \log_{1/2}\left(\frac{112}{200}\right) = \frac{180}{t_{1/2}} \] Now, multiply both sides by \(t_{1/2}\) and divide both sides by \(\log_{1/2}\left(\frac{112}{200}\right)\) to isolate the half-life: \[ t_{1/2} = \frac{180}{\log_{1/2}\left(\frac{112}{200}\right)} \] Finally, use a calculator to evaluate the expression, which gives: \[ t_{1/2} \approx 324.84 \: minutes \] The half-life of the unknown radioactive substance is approximately 324.84 minutes.

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Most popular questions from this chapter

Methyl acetate \(\left(\mathrm{CH}_{3} \mathrm{COOCH}_{3}\right)\) is formed by the reaction of acetic acid with methyl alcohol. If the methyl alcohol is labeled with oxygen-18, the oxygen-18 ends up in the methyl acetate: (a) Do the \(\mathrm{C}-\mathrm{OH}\) bond of the acid and the \(\mathrm{O}-\mathrm{H}\) bond of the alcohol break in the reaction, or do the \(\mathrm{O}-\mathrm{H}\) bond of the acid and the \(\mathrm{C}-\mathrm{OH}\) bond of the alcohol break? (b) Imagine a similar experiment using the radioisotope \({ }^{3} \mathrm{H}\), which is called tritium and is usually denoted T. Would the reaction between \(\mathrm{CH}_{3} \mathrm{COOH}\) and \(\mathrm{TOCH}_{3}\) provide the same information about which bond is broken as does the above experiment with \(\mathrm{H}^{18} \mathrm{OCH}_{3}\) ?

What do these symbols stand for? (a) \({ }_{0}^{0} \gamma,(\mathbf{b}){ }_{2}^{4} \mathrm{He},\) (c) \({ }_{0}^{1} \mathrm{n} .\)

In 1930 the American physicist Ernest Lawrence designed the first cyclotron in Berkeley, California. In 1937 Lawrence bombarded a molybdenum target with deuterium ions, producing for the first time an element not found in nature. What was this element? Starting with molybdenum-96 as your reactant, write a nuclear equation to represent this process.

How much time is required for a \(5.00-g\) sample of \({ }^{233}\) Pa to decay to \(0.625 \mathrm{~g}\) if the half-life for the beta decay of ${ }^{233} \mathrm{~Pa}$ is 27.4 days?

The energy from solar radiation falling on Earth is $1.07 \times 10^{16} \mathrm{~kJ} / \mathrm{min} .$ (a) How much loss of mass from the Sun occurs in one day from just the energy falling on Earth? (b) If the energy released in the reaction $$ { }^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{56}^{141} \mathrm{Ba}+{ }_{36}^{92} \mathrm{Kr}+3{ }_{0}^{1} \mathrm{n} $$ \(\left({ }^{235} \mathrm{U}\right.\) nuclear mass, \(234.9935 \mathrm{u} ;{ }^{141} \mathrm{Ba}\) nuclear mass, $140.8833 \mathrm{u} ;{ }^{92} \mathrm{Kr}\( nuclear mass, \)91.9021 \mathrm{u}$ ) is taken as typical of that occurring in a nuclear reactor, what mass of uranium- 235 is required to equal \(0.10 \%\) of the solar energy that falls on Earth in 1.0 day?
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