91Ó°ÊÓ

The term kinetic energy may be defined as the energy by virtue of its motion.

a)

When p and$\overline{p}$ are initially at rest, the energy can be calculated by the relation:

$$\begin{gathered} E_r=m_p{c}^2 \\ {E}_r=\left(1.67×{10}^{−27}\mathrm{kg}\right){\left(3.0×{10}^8\mathrm{m}/\mathrm{s}\right)}^2 \\ {E}_r=\left(1.503×{10}^{−10}\mathrm{J}\right)\left(\frac{1\mathrm{MeV}}{1.60×{10}^{−13}\mathrm{J}}\right) \\ {E}_r=939.375\mathrm{MeV} \end{gathered}$$

Now frequency:

$$\begin{gathered} f=\frac{E_r}{h} \\ f=\frac{1.503×{10}^{−10}\mathrm{J}}{6.626×{10}^{−34}\mathrm{J}â‹\dots \mathrm{s}} \\ f=2.27×{10}^{23}\mathrm{Hz} \end{gathered}$$

And wavelength:

$$\begin{gathered} \mathrm{λ}=\frac{c}{f} \\ \mathrm{λ}=\frac{3×{10}^8\mathrm{m}/\mathrm{s}}{2.27×{10}^{23}\mathrm{Hz}} \\ \mathrm{λ}=1.32\mathrm{fm} \end{gathered}$$

Hence, when and are initially at rest, the energy, frequency, and wavelength are $E=939.375 MeV, f=2.27×{10}^{23} Hz, \mathrm{λ}=1.32 fm$.

b)

Now when the p and$\overline{p}$ collide head-on, each with an initial kinetic energy of 620 MeV.

The energy can be calculated by the relation:

$$\begin{gathered} E_{\mathrm{γ}}=E_r+K \\ {E}_{\mathrm{γ}}=939.375\mathrm{MeV}+620\mathrm{MeV} \\ {E}_{\mathrm{γ}}=1559\mathrm{MeV} \end{gathered}$$

Frequency can be calculated as:

$$\begin{gathered} f=\frac{E_{\mathrm{γ}}}{h} \\ f=\frac{1559×{10}^6\mathrm{eV}}{4.136×{10}^{−15}\mathrm{eV}â‹\dots \mathrm{s}} \\ f=3.77×{10}^{23}\mathrm{Hz} \end{gathered}$$

And the wavelength is:

$$\begin{gathered} \mathrm{λ}=\frac{c}{f} \\ \mathrm{λ}=\frac{3×{10}^8\mathrm{m}/\mathrm{s}}{3.77×{10}^{23}{\mathrm{s}}^{−1}} \\ \mathrm{λ}=7.96×{10}^{−16}\mathrm{m} \end{gathered}$$

Hence, when the P and $\overline{p}$collide head-on, each with an initial kinetic energy of 620 MeV the $E=1559 MeV, f=3.77×{10}^{23} Hz, \mathrm{λ}=0.796 fm$.