Question 31.2: Detecting Proton Decay Measurements taken at the Super Kamio...
Detecting Proton Decay
Measurements taken at the Super Kamiokande neutrino detection facility in Japan (Fig. 31.6) indicate that the half-life of protons is at least 10³³ years.
A Estimate how long we would have to watch, on average, to see a proton in a glass of water decay.
B The Super Kamiokande neutrino facility contains 50 000 metric tons of water. Estimate the average time interval between detected proton decays in this much water if the half-life of a proton is 10³³ yr.
A:
To conceptualize the problem, imagine the number of protons in a glass of water. Although this number is huge, we know that the probability of a single proton undergoing decay is small, so we would expect to wait a long time before observing a decay. Because a half-life is provided in the problem, we categorize this problem as one in which we can apply our statistical analysis techniques from Section 30.3. To analyze the problem, let us estimate that a glass contains about 250 g of water. The number of molecules of water is
\begin{aligned}\frac{(250 g )\left(6.02 \times 10^{23} molecules / mol \right)}{18 g / mol } \\=8.4 \times 10^{24} molecules\end{aligned}
Each water molecule contains one proton in each of its two hydrogen atoms plus eight protons in its oxygen atom, for a total of ten. Therefore, 8.4 \times 10^{25} protons are in the glass of water. The decay constant is given by Equation 30.8:
\lambda=\frac{0.693}{T_{1 / 2}}=\frac{0.693}{10^{33} yr }=6.9 \times 10^{-34} yr ^{-1}
T_{1 / 2}=\frac{\ln 2}{\lambda}=\frac{0.693}{\lambda} [30.8]
This result is the probability that any one proton will decay in a year. The probability that any proton in our glass of water will decay in the one-year interval is (Eqs. 30.5 and 30.7)
R=\left(8.4 \times 10^{25}\right)\left(6.9 \times 10^{-34} yr ^{-1}\right)=5.8 \times 10^{-8} yr ^{-1}
\frac{d N}{d t}=-\lambda N [30.5]
R=\left|\frac{d N}{d t}\right|=N_{0} \lambda e^{-\lambda t}=R_{0} e^{-\lambda t} [30.7]
To finalize this part of the problem, note that we have to watch our glass of water for 1/R ≈ 17 million years! This answer is indeed a long time, as we suspected.
B:
We find the ratio of the number of molecules in 50 000 metric tons of water to that in the glass of water in part A, which will be same as the ratio of masses:
\begin{aligned}\frac{N_{\text {Kamiokande }}}{N_{\text {glass }}} &=\frac{m_{\text {Kamiokande }}}{m_{\text {glass }}} \\&=\frac{50000 \text { metric ton }}{250 g }\left(\frac{1000 kg }{1 \text { metric ton }}\right)\left(\frac{1000 g }{1 kg }\right) \\&=2.0 \times 10^{8}\end{aligned}
\begin{aligned}N_{\text {Kamiokande }} &=\left(2.0 \times 10^{8}\right) N_{\text {glass }} \\&=\left(2.0 \times 10^{8}\right)\left(8.4 \times 10^{24} \text { molecules }\right) \\&=1.7 \times 10^{33} \text { molecules }\end{aligned}
Each of these molecules contains ten protons. The probability that one of these protons will decay in one year is
R=(10)\left(1.7 \times 10^{33}\right)\left(6.9 \times 10^{-34} yr ^{-1}\right) \approx 12 yr ^{-1}
To finalize this part of the problem, note that the average time interval between decays is about one twelfth of a year, or approximately one month. This result is much shorter than the time interval in part A due to the tremendous amount of water in the detector facility.
Physics Now™ Practice the statistics of proton decay by logging into PhysicsNow at www.pop4e.com and going to Interactive Example 31.2.