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Chapter 21: Problem 35

Some watch dials are coated with a phosphor, like ZnS, and a polymer in which some of the \({ }^{1} \mathrm{H}\) atoms have been replaced by ${ }^{3} \mathrm{H}$ atoms, tritium. The phosphor emits light when struck by the beta particle from the tritium decay, causing the dials to glow in the dark. The half-life of tritium is 12.3 yr. If the light given off is assumed to be directly proportional to the amount of tritium, by how much will a dial be dimmed in a watch that is 50 yr old?

Short Answer

Expert verified
After 50 years, the watch dial coated with a phosphor like ZnS and tritium will be dimmed by approximately 94.3%, as the remaining tritium fraction is about 5.7%.

Step by step solution

01

Identify the given information and the formula to use

We know the half-life of tritium is 12.3 years and the watch is 50 years old. We also know that the light emitted is directly proportional to the amount of tritium. We can use the half-life formula to find the remaining percentage of tritium: \[ RemainingFraction = \frac{1}{2^{\frac{Time}{Half-Life}}}\]
02

Insert the known values into the formula

Now we shall plug the values into the formula. \[ RemainingFraction = \frac{1}{2^{\frac{50}{12.3}}}\]
03

Calculate the remaining tritium fraction

Next, we need to compute the remaining fraction of tritium: \[ RemainingFraction = \frac{1}{2^{\frac{50}{12.3}}} \approx 0.057\] This means that after 50 years, approximately 5.7% of the tritium remains.
04

Determine the dimming percentage

Since the light emitted is directly proportional to the amount of tritium, we can say that the remaining light emitted after 50 years is also 5.7%. Thus, to calculate how much the dial will be dimmed after 50 years, we need to subtract the remaining light percentage from 100%: \[ DimmingPercentage = 100\% - 5.7\% = 94.3\%\]
05

Interpret the result

After 50 years, the watch dial will be dimmed by approximately 94.3%.

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Most popular questions from this chapter

Write balanced equations for each of the following nuclear reactions: $(\mathbf{a}){ }_{92}^{238} \mathrm{U}(\mathrm{n}, \gamma){ }^{239} \mathrm{U},(\mathbf{b}){ }_{82}^{16} \mathrm{O}(\mathrm{p}, \alpha){ }^{13} \mathrm{~N},\( (c) \){ }_{8}^{18} \mathrm{O}\left(\mathrm{n}, \beta^{-}\right){ }^{19} \mathrm{~F}$.

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