Suggested languages for you:

Americas

Europe

Q14P

Expert-verified
Found in: Page 1364

### Fundamentals Of Physics

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

# Calculate the disintegration energy of the reactions$\left(a\right){{\mathbit{\pi }}}^{{\mathbf{+}}}{\mathbf{+}}{\mathbit{p}}{\mathbf{\to }}\mathbf{\sum }^{\mathbf{+}}{\mathbf{+}}{\mathbf{\text{}}}{{\mathbit{K}}}^{{\mathbf{+}}}{\mathbf{\text{and}}}\left(b\right){{\mathbit{K}}}^{{\mathbf{-}}}{\mathbf{+}}{\mathbit{p}}{\mathbf{\to }}{{\mathbit{\Lambda }}}^{{\mathbf{0}}}{\mathbf{+}}{{\mathbit{\pi }}}^{{\mathbf{0}}}$

(a) The value ${\pi }^{+}+p\to \sum ^{+}+\text{}{K}^{+}$ is $605.2\mathrm{Mev}$.

(b)The value ${K}^{-}+p\to {\Lambda }^{0}+{\pi }^{0}$ is $-181.4\mathrm{Mev}$.

See the step by step solution

## Step 1: Disintegration energy

Disintegration energy is lost by the nucleus undergoing radioactive decay. During radioactive decay, an element loses mass and energy by emitting radiation and ionizing particles. During this process of radioactive decay. an atom is transformed into another element.

The chemical species undergoing radioactive decay is called the parent nuclide, and the new chemical species is called the daughter nuclide.

.$Q=-\Delta m.{c}^{2}$

## Step 2 (a) : Solve π++p→∑++ K+

(1) Explanation

${m}_{{\pi }^{+}}\to 139.6\mathrm{MeV}\phantom{\rule{0ex}{0ex}}{m}_{p}\to 938.3\mathrm{MeV}\phantom{\rule{0ex}{0ex}}{m}_{\sum ^{+}}\to 1189.4\mathrm{MeV}\phantom{\rule{0ex}{0ex}}{m}_{{K}^{+}}\to 493.7\mathrm{MeV}$

(2) Solution

$Q=-\Delta m.{c}^{2}$ ….. (1)

Put the value in equation (1)

$\begin{array}{rcl}Q& =& \left({m}_{\sum ^{+}}+{m}_{{K}^{+}}-{m}_{{\pi }^{+}}-{m}_{p}\right){c}^{2}\\ & =& \left(1189.4+493.7-139.6-938.3\right)\mathrm{MeV}\\ & =& \left(1683.1-1077.9\right)\mathrm{MeV}\\ & =& 605.2\mathrm{Mev}\end{array}$

Hence the value ${\pi }^{+}+p\to \sum ^{+}+\text{}{K}^{+}$ is $605.2\mathrm{Mev}$.

## Step 3 (b): Solve K-+p→Λ0+π0

(1) Explanation

${m}_{{K}^{-}}\to 493.7\mathrm{MeV}\phantom{\rule{0ex}{0ex}}{m}_{p}\to 938.3\mathrm{MeV}\phantom{\rule{0ex}{0ex}}{m}_{{\Lambda }^{0}}\to 1115.6\mathrm{Mev}\phantom{\rule{0ex}{0ex}}{m}_{{\pi }^{0}}\to 135.0$

(2) Solution.

$\begin{array}{rcl}Q& =& -\Delta m.{c}^{2}\\ Q& =& \left({m}_{{\Lambda }^{0}}+{m}_{{\pi }^{0}}-{m}_{{K}^{-}}-{m}_{p}\right){c}^{2}\\ & =& \left(1115.6+135.0-493.7-938.3\right)\mathrm{MeV}\\ & =& \left(1250.6-1432\right)\mathrm{MeV}\\ & =& -181.4\mathrm{Mev}\end{array}$

Hence, the value ${K}^{-}+p\to {\Lambda }^{0}+{\pi }^{0}$is $-181.4\mathrm{Mev}$.