Question

In a reference frame where a ∆+ particle is at rest it decays into a proton and a high-energy photon (a “gamma ray”): ${\mathbf{∆}}^{\mathbf{+}}\mathbf{\to }{\mathit{P}}^{\mathbf{+}}\mathbf{+}{\mathit{\gamma }}^{\mathbf{+}}$. The mass of the ∆+ particle is 1232 $\mathbf{MeV}\mathbf{/}{\mathbf{c}}^{\mathbf{2}}$and the mass of the proton is 938 $\mathbf{MeV}\mathbf{/}{\mathbf{c}}^{\mathbf{2}}$(1 MeV = ${\mathbf{10}}^{\mathbf{6}}$eV). Calculate the energy of the gamma ray and the speed of the proton.

Short answer

The energy of gamma ray is $258.92\mathrm{MeV}$and the speed of the proton is$8.28\times {10}^7\mathrm{m}/\mathrm{s}$

Step-by-step solution

Identification of given data

The given data can be listed below,

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

Concept/Significance of high energy photon

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.

Determination of the energy of the gamma ray and the speed of the proton

The decay equation is given by,

${∆}^{+}\to p^{+}+\gamma$

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

From conservation of momentum is given by,

$P_{∆+}=P+P_{\gamma }$

Momentum of ${∆}^{+}$is given by,

$P_{∆+}=0$

Momentum of the photon is given by,

$P_{\gamma }=\frac{E}{c}$

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

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,

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

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,

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

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