Chapter 41, Problem 42P

(a)

To determine

The total energy of the decay products.

Explanation of Solution

Given: The given reaction is ${\Sigma }^0\to {\Lambda }^0+\gamma$ .

Formula used:

Write the expression for total energy of the decay products.

  $E=\left(m_{{\Sigma }^0}\right)c^2$ …… (1)

Here, $E$ is total energy, $m_{{\Sigma }^0}$ is rest mass of particle and $c$ is the speed of light.

Calculation:

Substitute $1193\text{MeV}/c^2$ for $m_{{\Sigma }^0}$ in equation (1).

  $\begin{array}{c}E=\left(1193\text{MeV}/c^2\right)c^2\\ =1193\text{MeV}\end{array}$

Conclusion:

Thus, the total energy of the decay products is $1193\text{MeV}$ .

(b)

To determine

The approximate momentum of the photon.

Explanation of Solution

Given: The given reaction is ${\Sigma }^0\to {\Lambda }^0+\gamma$ .

Formula used:

Write the expression for total energy of the decay products.

  $E=\left(m_{{\Sigma }^0}\right)c^2$

Here, $E$ is total energy, $m_{{\Sigma }^0}$ is rest mass of particle and $c$ is the speed of light.

Write the expression for the rest energy of baryon.

  $E_{{\Lambda }^0}=\left(m_{{\Lambda }^0}\right)c^2$

Here, $m_{{\Lambda }^0}$ is mass of baryon and $E_{{\Lambda }^0}$ is rest energy of baryon.

Write the expression for the rest energy of photon.

  $E_{\gamma }=E-E_{{\Lambda }^0}$

Substitute $\left(m_{{\Lambda }^0}\right)c^2$ for $E_{{\Lambda }^0}$ in above expression.

  $E_{\gamma }=E-\left(m_{{\Lambda }^0}\right)c^2$

Here, $E_{\gamma }$ is rest energy of photon.

Write the expression for momentum of photon.

  $p_{\gamma }=\frac{E_{\gamma }}{c}$

Substitute $E-\left(m_{{\Lambda }^0}\right)c^2$ for $E_{\gamma }$ in above expression,

  $p_{\gamma }=\frac{E-\left(m_{{\Lambda }^0}\right)c^2}{c}$ …… (2)

Here, $p_{\gamma }$ is momentum of photon.

Calculation:

Substitute $1193\text{MeV}$ for $E$ and $1116\text{MeV}/c^2$ for $m_{\Lambda }$ in equation (2).

  $\begin{array}{c}{p}_{\gamma }=\frac{1193\text{MeV}-\left(1116\text{MeV}/c^2\right)c^2}{c}\\ =77\text{MeV}/c\end{array}$

Conclusion:

Thus, the approximate momentum of the photon is $77\text{MeV}/c$ .

(c)

To determine

The kinetic energy of baryon.

Explanation of Solution

Given: The given reaction is ${\Sigma }^0\to {\Lambda }^0+\gamma$ .

Formula used:

Write the expression for total energy of the decay products.

  $E=\left(m_{{\Sigma }^0}\right)c^2$

Here, $E$ is total energy, $m_{{\Sigma }^0}$ is rest mass of particle and $c$ is the speed of light.

Write the expression for the rest energy of baryon.

  $E_{{\Lambda }^0}=\left(m_{{\Lambda }^0}\right)c^2$

Here, $m_{{\Lambda }^0}$ is mass of baryon and $E_{{\Lambda }^0}$ is rest energy of baryon.

Write the expression for the rest energy of photon.

  $E_{\gamma }=E-E_{{\Lambda }^0}$

Substitute $\left(m_{{\Lambda }^0}\right)c^2$ for $E_{{\Lambda }^0}$ in above expression.

  $E_{\gamma }=E-\left(m_{{\Lambda }^0}\right)c^2$

Here, $E_{\gamma }$ is rest energy of photon.

Write the expression for momentum of photon.

  $p_{\gamma }=\frac{E_{\gamma }}{c}$

Substitute $E-\left(m_{{\Lambda }^0}\right)c^2$ for $E_{\gamma }$ in above expression,

  $p_{\gamma }=\frac{E-\left(m_{{\Lambda }^0}\right)c^2}{c}$

Here, $p_{\gamma }$ is momentum of photon.

The momentum of baryon and photon is equal.

Write the expression for the kinetic energy of baryon.

  $K_{{\Lambda }^0}=\frac{p_{{\Lambda }^0}^2}{2m_{{\Lambda }^0}}$

Here, $p_{{\Lambda }^0}$ is momentum of baryon and $K_{{\Lambda }^0}$ is kinetic energy of baryon.

Substitute $p_{\gamma }$ for $p_{{\Lambda }^0}$ in above expression.

  $K_{{\Lambda }^0}=\frac{p_{\gamma }^2}{2m_{{\Lambda }^0}}$ …… (3)

Calculation:

Substitute $77\text{MeV}/c$ for $p_{\gamma }$ and $1116\text{MeV}/c^2$ for $m_{\Lambda }$ in equation (3).

  $\begin{array}{c}{K}_{{\Lambda }^0}=\frac{{\left(77\text{MeV}/c\right)}^2}{2\left(1116\text{MeV}/c^2\right)}\\ =2.7\text{MeV}\end{array}$

Conclusion:

Thus, the kinetic energy of baryon is $2.7\text{MeV}$ .

(d)

To determine

The estimate energy of the photon; the estimate momentum of the photon.

Explanation of Solution

Given: The given reaction is ${\Sigma }^0\to {\Lambda }^0+\gamma$ .

Formula used:

Write the expression for total energy of the decay products.

  $E=\left(m_{{\Sigma }^0}\right)c^2$

Here, $E$ is total energy, $m_{{\Sigma }^0}$ is rest mass of particle and $c$ is the speed of light.

Write the expression for the rest energy of baryon.

  $E_{{\Lambda }^0}=\left(m_{{\Lambda }^0}\right)c^2$

Here, $m_{{\Lambda }^0}$ is mass of baryon and $E_{{\Lambda }^0}$ is rest energy of baryon.

Write the expression for the rest energy of photon.

  $E_{\gamma }=E-E_{{\Lambda }^0}$

Substitute $\left(m_{{\Lambda }^0}\right)c^2$ for $E_{{\Lambda }^0}$ in above expression.

  $E_{\gamma }=E-\left(m_{{\Lambda }^0}\right)c^2$

Here, $E_{\gamma }$ is rest energy of photon.

Write the expression for momentum of photon.

  $p_{\gamma }=\frac{E_{\gamma }}{c}$

Substitute $E-\left(m_{{\Lambda }^0}\right)c^2$ for $E_{\gamma }$ in above expression,

  $p_{\gamma }=\frac{E-\left(m_{{\Lambda }^0}\right)c^2}{c}$

Here, $p_{\gamma }$ is momentum of photon.

The momentum of baryon and photon is equal.

Write the expression for the kinetic energy of baryon.

  $K_{{\Lambda }^0}=\frac{p_{{\Lambda }^0}^2}{2m_{{\Lambda }^0}}$

Here, $p_{{\Lambda }^0}$ is momentum of baryon and $K_{{\Lambda }^0}$ is kinetic energy of baryon.

Substitute $p_{\gamma }$ for $p_{{\Lambda }^0}$ in above expression.

  $K_{{\Lambda }^0}=\frac{p_{\gamma }^2}{2m_{{\Lambda }^0}}$

Write the expression for estimate energy of photon.

  ${E^{\prime }}_{\gamma }=E-\left(m_{{\Lambda }^0}\right)c^2-K_{{\Lambda }^0}$ …… (4)

Here, ${E^{\prime }}_{\gamma }$ is estimate energy of photon.

Write the expression for estimate momentum of photon.

  ${p^{\prime }}_{\gamma }=\frac{{E^{\prime }}_{\gamma }}{c}$ …… (5)

Here, ${p^{\prime }}_{\gamma }$ is estimate momentum of photon.

Calculation:

Substitute $1193\text{MeV}$ for $E$ , $1116\text{MeV}/c^2$ for $m_{\Lambda }$ and $2.7\text{MeV}$ for $K_{{\Lambda }^0}$ in equation (4).

  $\begin{array}{c}{E^{\prime }}_{\gamma }=1193\text{MeV}-\left(1116\text{MeV}/c^2\right)c^2-2.7\text{MeV}\\ =74\text{MeV}\end{array}$

Substitute $74\text{MeV}$ for ${E^{\prime }}_{\gamma }$ in equation (5).

  ${p^{\prime }}_{\gamma }=74\frac{\text{MeV}}{c}$

Conclusion:

The estimate energy of the photon is $74\text{MeV}$ ; the estimate momentum of the photon is $74\text{MeV}/c$ .