Chapter 30.3, Problem 40PP

To determine

To calculate:

The amount of energy released when the mass of muon decays into one electron and two neutrons.

Answer to Problem 40PP

The amount of energy released is $105\text{MeV}$ .

Explanation of Solution

Given:

For a muon,

Mass of muon $=0.1135\text{u}$

Mass of an electron $=$$0.00054858\text{u}$

Mass of a neutrino $=\text{3}\text{.435328}e\text{-}\text{10}\text{u}$

Formula Used:

Total Energy $E=m{c}^2$

Where,

  $m$ : Mass

  $c$: Speed of light

  $E$ : Total Energy

Calculation:

According to the given condition,

Mass of muon decays into one electron and two neutrons.

First find $\Delta m$

  $\Delta m=0.1135-0.0005848-2\times \left(3.435328e-10\right)=0.11295\text{u}$

Now to find the total energy released, by using the formula, $E=m{c}^2$

Substituting all the values in the given formula,

  $\begin{array}{l}=\left(0.11295\text{u}\right)\times \left(1.66\times {10}^{-27}\frac{\text{Kg}}{\text{u}}\right)\times {\left(3.10\times {10}^8\right)}^2\\ =1.68747\times {10}^{-11}\text{J}\end{array}$

To convert into $\text{MeV}$ , multiplying Energy by $\left(6.242e+12\right)$

  $=\left(6.242e+12\right)\times \left(.68747\times {10}^{-11}\text{J}\right)$

  $=105\text{MeV}$

Conclusion:

The amount of energy released when the mass of muon decays into one electron and two neutrons is $105\text{MeV}$ .

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