Suggested languages for you:

Americas

Europe

Q23P

Expert-verified
Found in: Page 1364

Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

Consider the decay ${{\mathbit{\Lambda }}}^{{\mathbf{0}}}{\mathbf{\to }}{\mathbit{p}}{\mathbf{+}}{{\mathbit{\pi }}}^{{\mathbf{-}}}$with the${{\mathbit{\Lambda }}}^{{\mathbf{0}}}$ at rest. (a) Calculate the disintegration energy. What is the kinetic energy of (b) the proton and (c) the pion?

(a) The disintegration energy is.$37.7\text{\hspace{0.17em}Mev}$

(b) The kinetic energy of proton is.$5.35\text{\hspace{0.17em}Mev}$

(c) The kinetic energy of pion is.$32.4\text{\hspace{0.17em}Mev}$

See the step by step solution

Step 1: (a) Evaluate disintegration energy.

The given reaction is written as:

${\Lambda }^{0}\to p+{\pi }^{-}$

Solve the disintegration energy as follows:

$\begin{array}{c}Q=\Delta m{c}^{2}\\ =\left({m}_{{\Lambda }^{0}}-{m}_{p}-{m}_{{\pi }^{-}}\right){c}^{2}\end{array}$

Substitute the values from the standard data.

$\begin{array}{c}Q=\left(1115.6-938.3-139.6\right)\text{\hspace{0.17em}}Mev\\ =37.7\text{\hspace{0.17em}}Mev\end{array}$

Thus, the disintegration energy is.$37.7\text{\hspace{0.17em}}Mev$

Step 2: (b) Evaluate the kinetic energy of proton.

Solve the kinetic energy of proton in the given reaction as follows;

${K}_{p}=\frac{{\left({E}_{{}_{\Lambda }}-{E}_{p}\right)}^{2}-{E}_{\pi }^{2}}{2{E}_{\Lambda }}$

Substitute the values from the standard data

$\begin{array}{c}{K}_{p}=\frac{{\left(1115.6-938.3\right)}^{2}-{\left(139.6\right)}^{2}}{2\left(1115.6\right)}\text{\hspace{0.17em}Mev}\\ =5.35\text{\hspace{0.17em}Mev}\end{array}$

Thus, the kinetic energy of proton is .$5.35\text{\hspace{0.17em}}Mev$

Step 3: (c) Evaluate the kinetic energy of pion.

Solve the kinetic energy of proton in the given reaction as follows;

$\begin{array}{c}{K}_{\pi }=Q-{K}_{p}\\ =\left(37.7-5.35\right)\text{\hspace{0.17em}Mev}\\ =32.4\text{​\hspace{0.17em}Mev}\end{array}$

Thus, the kinetic energy of pion is.$32.4\text{\hspace{0.17em}}Mev$