Question

(a) What is the uncertainty in the energy released in the decay of a \(\tau {\rm{ }} - \)due to its short lifetime?

(b) Is the uncertainty in this energy greater than or less than the uncertainty in the mass of the tau neutrino? Discuss the source of the uncertainty.

Short answer

a) The energy released in the decay is\({\rm{\Delta E = 1}}{\rm{.812 \times 1}}{{\rm{0}}^{{\rm{22}}}}{\rm{j}}\).

b)It only applies to the \({{\rm{\mu }}^{\rm{ - }}}\)because neutrinos have no rest mass energy or, as we showed, it is almost zero.

Step-by-step solution

Concept Introduction

Heisenberg’s uncertainty principle states that the energy content of a particle and the time interval cannot be simultaneously measured accurately. Always there will be an uncertainty in their value, which is related as-

\({\bf{\Delta E}}{\bf{.\Delta T}} \le \frac{{\bf{h}}}{{{\bf{4\pi }}}}\)

Calculating the energy

a)

According to the Heisenberg uncertainty principle, energy uncertainty is connected to temporal uncertainty in the following way:

\(\Delta E\Delta t \simeq \frac{h}{{4\pi }}\)

Where h is Planck's constant, which equals\(h = 6.626 \times {10^{ - 34}}\;{\rm{J}}{\rm{.s}}\)

The lifespan for\({\tau ^ - }\)may be found in table\({\rm{33}}{\rm{.2}}\).

\(\Delta {T_t} = 2.91 \times {10^{ - 13}}\;{\rm{s}}\)

Therefore, to applying the Heisenberg uncertainty principle to the uncertainty, we obtain:

\(\begin{array}{c}\Delta {E_\tau } \simeq \frac{h}{{4\pi \Delta {T_\tau }}}\\ \simeq \frac{{\left( {6.626 \times {{10}^{ - 34}}\;{\rm{J}}{\rm{.s}}} \right)}}{{4\pi \times \left( {2.91 \times {{10}^{ - 13}}\;{\rm{s}}} \right)}}\\ \simeq 1.812 \times {10^{22}}\;{\rm{J}}\end{array}\)

Therefore, the energy released in the decay is \(\Delta E = 1.812 \times {10^{22}}\;{\rm{J}}\).

Explanation

b)

In this section, we wished to compare the energy released in the decay of\({\tau ^ - }\)to the mass uncertainty for the tau neutrino. As a result,\({\tau ^ - }\)depreciates in the following way:

\({\tau ^ - } \to {\mu ^ - } + {\bar \nu _\mu } + {\nu _\tau }.\)

Now\({\rm{0}}\;\left( {{\rm{ < 31}}\;\;{\rm{Mev/}}{{\rm{c}}^{\rm{2}}}} \right)\)is the uncertainty in the tau neutrino from table\({\rm{33}}{\rm{.2}}{\rm{.}}\)

As a result, even if we use the greatest number for the mass, the energy released in will be substantially lower.

The two additional particles, \({{\rm{\mu }}^{\rm{ - }}}{\rm{ and }}{{\rm{\nu }}_{\rm{\mu }}}\), are also sources of uncertainty in the energy from decay. Actually, it only applies to the \({{\rm{\mu }}^{\rm{ - }}}\), because neutrinos have no rest mass energy or, as we have shown, it is almost zero.