Suppose you are designing a proton decay experiment and you can detect \({\rm{50}}\) percent of the proton decays in a tank of water.

(a) How many kilograms of water would you need to see one decay per month, assuming a lifetime of \({\rm{1}}{{\rm{0}}^{{\rm{31}}}}{\rm{ y}}\)?

(b) How many cubic meters of water is this?

(c) If the actual lifetime is \({\rm{1}}{{\rm{0}}^{{\rm{33}}}}{\rm{ y}}\), how long would you have to wait on an average to see a single proton decay?

Density of the liquid is given by,

\(\rho = \frac{m}{V}\)

Here\(\rho \)is the density,\(m\)is the mass and\(V\)is the volume.

(a)

Plain water is consisted out of\({\rm{2}}\)hydrogen atoms and\({\rm{1}}\)oxygen atom. A hydrogen atom contains\({\rm{1}}\)proton, oxygen contains\({\rm{8}}\)so this gives a net of\({\rm{10}}\)protons per hydrogen molecule. Molar mass of water is\({M_u} = {\rm{18}}{\rm{.02 g/mol}} = {\rm{18}}{\rm{.02 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{ kg/mol}}\). Mean lifetime is given with\(\tau = {\rm{1}}{{\rm{0}}^{{\rm{31}}}}{\rm{y}} = {\rm{12 \times 1}}{{\rm{0}}^{{\rm{31}}}}\). The number of particles is

\(N(t) = {N_0}{e^{ - \frac{t}{\tau }}}\)

As on average, one proton decays in every\({10^{31}}\)years which is equivalent to\(12 \times {10^{31}}\)months.

So, for one decay every months,

\(N = 1.2 \times {10^{32}}\,{\rm{protons}}\)

Here\(N\)is the number of protons.

Since, only\(50\% \)of the actual decays are detected. So, it needs twice of the number of protons to observe in single decay per month.

Therefore,

\(N = 2.4 \times {10^{32}}\,{\rm{protons}}\)

A hydrogen atom contains\({\rm{1}}\)proton, oxygen contains\({\rm{8}}\)so this gives a net of\({\rm{10}}\)protons per hydrogen molecule.

\(\begin{aligned}{}{N_{molecule}}\left( {{{\rm{H}}_{\rm{2}}}{\rm{O}}} \right) &= \frac{{2.4 \times {{10}^{32}}}}{{10}}\\ &= 2.4 \times {10^{31}}\end{aligned}\)

The calculation expressed in mole

\(\begin{aligned}{}{N_0} &= \left( {{\rm{24 \times 1}}{{\rm{0}}^{{\rm{31}}}}} \right){\rm{/}}\left( {{\rm{6}}{\rm{.022 \times 1}}{{\rm{0}}^{{\rm{23}}}}{\rm{\;mo}}{{\rm{l}}^{{\rm{ - 1}}}}} \right)\\ &= {\rm{3}}{\rm{.99 \times 1}}{{\rm{0}}^{\rm{8}}}{\rm{\;mol}}\end{aligned}\)

Every molecule of water contains\({\rm{10}}\)protons so,

\(\begin{aligned}{}{N_w} &= {{\rm{N}}_{\rm{0}}}{\rm{/10}}\\ &= \left( {{\rm{3}}{\rm{.99 \times 1}}{{\rm{0}}^{\rm{8}}}{\rm{\;mol}}} \right){\rm{/10}}\\ &= {\rm{3}}{\rm{.99 \times 1}}{{\rm{0}}^{\rm{7}}}{\rm{\;mol}}\end{aligned}\)

Finally, the total mass required is obtained by using\({{\rm{M}}_{\rm{u}}}\)–

\(\begin{aligned}{}m &= {M_u}{N_w}\\ &= \left( {{\rm{18}}{\rm{.02 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{\;kg/mol}}} \right)\left( {{\rm{3}}{\rm{.99 \times 1}}{{\rm{0}}^{\rm{7}}}{\rm{\;mol}}} \right)\\ &= {\rm{7}}{\rm{.19 \times 1}}{{\rm{0}}^{\rm{5}}}{\rm{\;kg}}\end{aligned}\)

Therefore, the value for mass of water is obtained as \(m = {\rm{7}}{\rm{.19 \times 1}}{{\rm{0}}^{\rm{5}}}{\rm{\;kg}}\).

(b)

Density of water is \(\rho = {\rm{1}}{{\rm{0}}^{\rm{3}}}\,{\rm{kg}}{{\rm{m}}^{{\rm{ - 3}}}}\), so water in cubic meters is calculated as –

\(\begin{aligned}{}V &= m/\rho \\ &= \left( {{\rm{7}}{\rm{.19 \times 1}}{{\rm{0}}^{\rm{5}}}{\rm{\;kg}}} \right){\rm{/}}\left( {{\rm{1000kg}}{{\rm{m}}^{{\rm{ - 3}}}}} \right)\\ &= {\rm{7}}{\rm{.19 \times 1}}{{\rm{0}}^{\rm{2}}}{\rm{\;}}{{\rm{m}}^{\rm{3}}}\end{aligned}\)

Therefore, the amount of water in cubic meters is obtained as \(V = {\rm{7}}{\rm{.19 \times 1}}{{\rm{0}}^{\rm{2}}}{\rm{\;}}{{\rm{m}}^{\rm{3}}}\).

(c)

Notice that \(\tau '{\rm{ = 1}}{{\rm{0}}^{\rm{2}}}{\rm{ \times }}\tau \) so a person would have to wait

\(\begin{aligned}{c}t' &= {\rm{1}}{{\rm{0}}^{\rm{2}}}{\rm{ \times (1 month )}}\\ &= {\rm{100 months}}\end{aligned}\)

Therefore, the value for time is obtained as \(t' = 100\,{\rm{months}}\).