91影视

The given data can be listed below,

• The mass of the Particle is, $m_{鈭�+}=1232\mathrm{MeV}/{\mathrm{c}}^2$.
• The mass of the proton is, $m_P=938\mathrm{MeV}/{\mathrm{c}}^2$.
• 1 Mega electron volt is equal to, 1 MeV = ${10}^6\mathrm{eV}$.

The High-energy photons include x-rays and gamma rays, which are associated with high-frequency radiation. The penetrating power of these photons is greater. The nucleus of radioactive atoms emits gamma rays.

The decay equation is given by,

${鈭�}^{+}鈫�p^{+}+纬$

Here,$p^{+}$is proton and $纬$is photon.

From conservation of momentum is given by,

$P_{鈭�+}=P+P_{纬}$

Momentum of ${鈭�}^{+}$is given by,

$P_{鈭�+}=0$

Momentum of the photon is given by,

$P_{纬}=\frac{E}{c}$

Here, E is the emitted energy of photon and c is the speed of light whose value is $3脳{10}^8\mathrm{m}/\mathrm{s}$.

Substitute all the values in the momentum equation.

$P=-\frac{E}{c}$

The magnitude of momentum of proton is given by,

$P=\frac{E}{c}$

From conservation law, energy is given by,

$$\begin{gathered} m_{鈭�+}{c}^2=\sqrt{{\left(P{c}^2\right)}^2+\left(m_p{c}^2\right)}+E \\ {m}_{鈭�+}{c}^2-E=\sqrt{{\left(P{c}^2\right)}^2+\left(m_p{c}^2\right)} \\ E=\frac{{\left(m_{鈭�+}{c}^2\right)}^2-{\left(m_p{c}^2\right)}^2}{2m_{鈭�+}{c}^2} \end{gathered}$$

Substitute values in the above,

$$\begin{gathered} E=\frac{{(1232\mathrm{MeV})}^2-{(938\mathrm{MeV})}^2}{2(1232\mathrm{MeV})} \\ =258.92\mathrm{MeV} \end{gathered}$$

Thus, the energy of gamma ray is 258.92 MeV.

The speed of the proton is given by,

$V_p=\frac{E}{m_{p}c}$

Here, $m_p$is the mass of the proton, E is the energy.

Substitute values in the above,

$$\begin{gathered} V_p=\frac{258.92\mathrm{MeV}}{(938\mathrm{MeV}/{\mathrm{c}}^2)}\left(\frac{3脳{10}^8\mathrm{m}/\mathrm{s}}{1\mathrm{c}}\right) \\ =8.28脳{10}^7\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the speed of the proton is $8.28脳{10}^7\mathrm{m}/\mathrm{s}$.