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Q14P

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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 $-181.4\mathrm{Mev}$.