Chapter 32, Problem 20P

To determine

(a) To Determine:

The law of conservation of linear momentum

Answer to Problem 20P

Solution:

To verify, ${\text{m}}_{{\text{Λ}}^{\text{0}}}{\text{c}}^{\text{2}}\ne {\text{m}}_n{\text{c}}^{\text{2}}{\text{+m}}_{{\text{π}}^{\text{-}}}{\text{c}}^{\text{2}}$.

Explanation of Solution

Given Info:

The mass of ${\text{Λ}}^{\text{o}}$ is 1115.7MeV/c².

The mass of ‘n’ is 939.36 MeV/c².

The mass of ‘ ${\text{π}}^{\text{-}}$’ is 139.6MeV/c².

Formula Used:

${\text{P}}_{{\Lambda }^0}{\text{=m}}_{{\text{Λ}}^{\text{0}}}{\text{c}}^{\text{2}}$

• ‘ ${\text{P}}_{{\Lambda }^0}$’ is the momentum of ${\text{Λ}}^{\text{o}}$.

• ‘ ${\text{m}}_{{\text{Λ}}^{\text{0}}}$’is mass of ${\text{Λ}}^{\text{o}}$.

• ‘ c’ is velocity of light.

${\text{P}}_{\text{n}}{\text{+P}}_{{\pi }^{-}}{\text{=m}}_n{\text{c}}^{\text{2}}{\text{+m}}_{{\text{π}}^{\text{-}}}{\text{c}}^{\text{2}}$

• ‘ ${\text{P}}_n$’ is the momentum of ‘n’.

• ‘ ${\text{P}}_{{\pi }^{-}}$’ is the momentum of ${\pi }^{-}$.

• ‘ ${\text{m}}_{\text{n}}$’is mass of ‘ $\text{n}$’.

• ‘ ${\text{m}}_{{\pi }^{-}}$’is mass of ‘ ${\pi }^{-}$’

• ‘ c’ is velocity of light

Calculation:

Substitute 1115.7MeV/c² for ${\text{m}}_{{\text{Λ}}^{\text{0}}}$, to find ${\text{P}}_{{\Lambda }^0}$

${\text{P}}_{{\Lambda }^0}\text{=}\left(\text{1115}{\text{.7MeV/c}}^{\text{2}}\right){\text{×c}}^{\text{2}}$

${\text{P}}_{{\Lambda }^0}=\text{1115}\text{.7MeV}$ (I)

Substitute 939.6 MeV/c² for ‘n’ and 139.6MeV/c² for ‘ ${\text{π}}^{\text{-}}$’ to find ‘ ${\text{P}}_{\text{n}}{\text{+P}}_{{\pi }^{-}}$’,

${\text{P}}_{\text{n}}{\text{+P}}_{{\pi }^{-}}\text{=}\left(\text{939}\text{.6}{\text{MeV/c}}^{\text{2}}\right){\text{×c}}^{\text{2}}\text{+}\left(\text{139}\text{.6}{\text{MeV/c}}^{\text{2}}\right){\text{×c}}^{\text{2}}$

${\text{P}}_{\text{n}}{\text{+P}}_{{\text{π}}^{\text{-}}}=\text{1079}\text{.2}\text{MeV}$ (II)

From equation (I)and (II):

${\text{m}}_{{\text{Λ}}^{\text{0}}}{\text{c}}^{\text{2}}\ne {\text{m}}_{\text{n}}{\text{c}}^{\text{2}}{\text{+m}}_{{\text{π}}^{\text{-}}}{\text{c}}^{\text{2}}$ (III)

To determine

(b) To Determine:

The law of conservation of linear momentum

Answer to Problem 20P

Solution:

To verify, ${\text{m}}_{{\text{Λ}}^{\text{0}}}{\text{c}}^{\text{2}}\ne {\text{m}}_{\text{p}}{\text{c}}^{\text{2}}{\text{+m}}_{K^{\text{-}}}{\text{c}}^{\text{2}}$.

Explanation of Solution

Given Info:

The mass of ${\text{Λ}}^{\text{o}}$ is 1115.7MeV/c².

The mass of ‘p’ is 938.3 MeV/c².

The mass of ${\text{K}}^{\text{-}}$ is 493.7 MeV/c².

Formula Used:

${\text{P}}_{{\Lambda }^0}{\text{=m}}_{{\text{Λ}}^{\text{0}}}{\text{c}}^{\text{2}}$

• ‘ ${\text{P}}_{{\Lambda }^0}$’ is the momentum of ${\text{Λ}}^{\text{o}}$.

• ‘ ${\text{m}}_{{\text{Λ}}^{\text{0}}}$’is mass of ${\text{Λ}}^{\text{o}}$.

• ‘ c’ is velocity of light.

${\text{P}}_{\text{p}}{\text{+P}}_{{\text{K}}^{-}}{\text{=m}}_{\text{p}}{\text{c}}^{\text{2}}{\text{+m}}_{{\text{K}}^{\text{-}}}{\text{c}}^{\text{2}}$

• ‘ ${\text{P}}_{\text{p}}$’ is the momentum of ‘p’.

• ‘ ${\text{P}}_{{\text{K}}^{-}}$’ is the momentum of ‘ ${\text{K}}^{-}$’.

• ‘ ${\text{m}}_{\text{P}}$’is mass of ‘p’

• ‘ ${\text{m}}_{{\text{K}}^{-}}$’is mass of ‘ ${\text{K}}^{\text{-}}$’.

• ‘ c’ is velocity of light

Calculation:

Substitute 1115.7MeV/c² for ${\text{m}}_{{\text{Λ}}^{\text{0}}}$, to find ${\text{P}}_{{\Lambda }^0}$

${\text{P}}_{{\Lambda }^0}\text{=}\left(\text{1115}{\text{.7MeV/c}}^{\text{2}}\right){\text{×c}}^{\text{2}}$

${\text{P}}_{{\Lambda }^0}=\text{1115}\text{.7MeV}$ (IV)

Substitute 938.3 MeV/c² for ‘n’ and 139.6MeV/c² for ${\text{π}}^{\text{-}}$ to find ${\text{P}}_{\text{p}}{\text{+P}}_{{\text{K}}^{-}}$

${\text{P}}_{\text{P}}{\text{+P}}_{{\text{K}}^{-}}\text{=}\left(\text{938}\text{.3}{\text{MeV/c}}^{\text{2}}\right){\text{×c}}^{\text{2}}\text{+}\left(493.7{\text{MeV/c}}^{\text{2}}\right){\text{×c}}^{\text{2}}$

${\text{P}}_{\text{p}}{\text{+P}}_{{\text{K}}^{\text{-}}}=\text{1432}\text{MeV}$ (V)

From equation (IV)and (V):

${\text{m}}_{{\text{Λ}}^{\text{0}}}{\text{c}}^{\text{2}}\ne {\text{m}}_{\text{p}}{\text{c}}^{\text{2}}{\text{+m}}_{{\text{K}}^{\text{-}}}{\text{c}}^{\text{2}}$ (VI)

Conclusion:

From equation (VI), the equation ${\text{Λ}}^{\text{o}}\bcancel{\to }{\text{n+K}}^{\text{-}}$ is not obeying principle of conservation of linear momentum. Therefore, ${\text{Λ}}^{\text{o}}$ cannot decay to ‘n’ and ${\text{K}}^{\text{-}}$

To determine

(c) To determine:

The law of conservation of linear momentum

Answer to Problem 20P

Solution:

To verify, $\begin{array}{l}{\text{m}}_{{\text{Λ}}^{\text{0}}}{\text{c}}^{\text{2}}\ne {\text{m}}_{{\pi }^{+}}{\text{c}}^{\text{2}}{\text{+m}}_{{\pi }^{\text{-}}}{\text{c}}^{\text{2}}\\ \end{array}$.

Explanation of Solution

Given Info:

The mass of ${\text{Λ}}^{\text{o}}$ is 1115.7 MeV/c².

The mass of ‘ ${\pi }^{+}$’ is 139.6 MeV/c².

The mass of ‘ ${\pi }^{\text{-}}$’ is 139.6 MeV/c².

Formula Used:

${\text{P}}_{{\Lambda }^0}{\text{=m}}_{{\text{Λ}}^{\text{0}}}{\text{c}}^{\text{2}}$

• ‘ ${\text{P}}_{{\Lambda }^0}$’ is the momentum of ${\text{Λ}}^{\text{o}}$.

• ‘ ${\text{m}}_{{\text{Λ}}^{\text{0}}}$’is mass of ${\text{Λ}}^{\text{o}}$.

• ‘ c’ is velocity of light.

${\text{P}}_{{\pi }^{+}}{\text{+P}}_{{\pi }^{-}}{\text{=m}}_{{\pi }^{+}}{\text{c}}^{\text{2}}{\text{+m}}_{{\text{π}}^{\text{-}}}{\text{c}}^{\text{2}}$

• ‘ ${\text{P}}_{{\pi }^{+}}$’ is the momentum of ‘ ${\pi }^{+}$’.

• ‘ ${\text{P}}_{{\pi }^{-}}$’ is the momentum of ‘ ${\pi }^{-}$’.

• ‘ ${\text{m}}_{{\pi }^{+}}$’is mass of ‘ ${\pi }^{+}$’

• ‘ ${\text{m}}_{{\pi }^{-}}$’is mass of ‘ ${\pi }^{\text{-}}$’.

• ‘ c’ is velocity of light

Calculation:

Substitute 1115.7MeV/c² for ${\text{m}}_{{\text{Λ}}^{\text{0}}}$, to find ${\text{P}}_{{\Lambda }^0}$

${\text{P}}_{{\Lambda }^0}\text{=}\left(\text{1115}{\text{.7MeV/c}}^{\text{2}}\right){\text{×c}}^{\text{2}}$

${\text{P}}_{{\Lambda }^0}=\text{1115}\text{.7MeV}$ (VII)

Substitute 139.6 MeV/c² for ‘ ${\pi }^{+}$’ and 139.6MeV/c² for ${\text{π}}^{\text{-}}$ to find ${\text{P}}_{{\pi }^{+}}{\text{+P}}_{{\pi }^{-}}$

${\text{P}}_{{\pi }^{+}}{\text{+P}}_{{\pi }^{-}}\text{=}\left(\text{139}{\text{.6MeV/c}}^{\text{2}}\right){\text{×c}}^{\text{2}}\text{+}\left(139.6{\text{MeV/c}}^{\text{2}}\right){\text{×c}}^{\text{2}}$

${\text{P}}_{{\pi }^{+}}{\text{+P}}_{{\pi }^{\text{-}}}=\text{279}\text{.2}\text{MeV}$ (VIII)

From equation (VII)and (VIII):

${\text{m}}_{{\text{Λ}}^{\text{0}}}{\text{c}}^{\text{2}}\ne {\text{m}}_{{\pi }^{+}}{\text{c}}^{\text{2}}{\text{+m}}_{{\pi }^{\text{-}}}{\text{c}}^{\text{2}}$ (IX)