91Ó°ÊÓ

The wavelength is described as the distance over the shape of a particular wave repeats. However, wavelength is described as the distance amongst the identical points of a waveform.

If the pair remains stationary, then the total amount of energy is in the form of the internal energy. Then the equation of the energy of the antiproton and the proton is expressed as:

$$\begin{gathered} \frac{hc}{\mathrm{λ}}=2m_p{c}^2 \\ \mathrm{λ}=\frac{hc}{2m_p{c}^2}âˆ\pounds \\ \mathrm{λ}=\frac{h}{2m_{p}C} \end{gathered}$$

Here, $\mathrm{λ}$is the wavelength, $h$is the Planck’s constant, $c$is the velocity of the light and $m_p$is the mass of the proton.

Substitute $6.626×{10}^{-34}\mathrm{J}·\mathrm{s}\text{for}h,1.67×{10}^{-27}\mathrm{kg}\text{for}{m}_p\text{and}3×{10}^8\mathrm{m}/\mathrm{s}$for $c$in the above equation.

$$\begin{gathered} \mathrm{λ}=\frac{6.626×{10}^{-34}\mathrm{J}·\mathrm{s}}{2\left(1.67×{10}^{-27}\mathrm{kg}\right)\left(3×{10}^8\mathrm{m}/\mathrm{s}\right)} \\ =\frac{6.626×{10}^{-34}\mathrm{J}·\mathrm{s}}{2\left(5.01×{10}^{-19}\mathrm{kg}·\mathrm{m}/\mathrm{s}\right)} \\ =\frac{6.626×{10}^{-34}\mathrm{J}·\mathrm{s}}{\left(1.002×{10}^{-18}\mathrm{kg}·\mathrm{m}/\mathrm{s}\right)} \\ =6.61×{10}^{-16}\mathrm{J}·{\mathrm{s}}^2/\mathrm{kg}·\mathrm{m} \end{gathered}$$

Hence, further as.

$$\begin{gathered} \mathrm{λ}=6.61×{10}^{-16}\mathrm{J}·{\mathrm{s}}^2/\mathrm{kg}·\mathrm{m}×\frac{\mathrm{kg}·{\mathrm{m}}^2/{\mathrm{s}}^2}{1\mathrm{J}} \\ =6.61×{10}^{-16}\mathrm{kg}·{\mathrm{m}}^2{\mathrm{s}}^2/\mathrm{kg}·\mathrm{m}·{\mathrm{s}}^2 \\ =6.61×{10}^{-16}\mathrm{m} \end{gathered}$$

Thus, the wavelength of the photon if the pair is stationary after creation is$=6.61×{10}^{-16}\mathrm{m}$ .

The equation of the wavelength of the proton is expressed as:

$$\begin{gathered} \frac{hc}{\mathrm{λ}}=2{\mathrm{γ}}_{0.6c}{m}_p{c}^2 \\ \mathrm{λ}=\frac{hc}{2{\mathrm{γ}}_{0.6c}{m}_p{c}^2} \\ \mathrm{λ}=\frac{hc}{2{\mathrm{γ}}_{0.6c}{m}_{p}c} \end{gathered}$$

Here, $\mathrm{λ}$is the wavelength, $h$is the Planck’s constant, localid="1658241714488" $c$is the velocity of the light,${\mathrm{γ}}_{0.6c}$is the velocity of the pair and localid="1657595455489" $m_p$is the mass of the proton.

Substitutelocalid="1657592042370" $6.626×{10}^{-34}\mathrm{J}·\mathrm{s}\text{for}h,1.67×{10}^{-27}\mathrm{kg}\text{for}{m}_p,1.25\text{for}{\mathrm{γ}}_{0.8c}\text{and}3×{10}^8\mathrm{m}/\mathrm{s}$in $\mathrm{for}c$the above equation.

localid="1657592117578" $$\begin{gathered} \mathrm{λ}=\frac{6.626×{10}^{-34}\mathrm{J}·\mathrm{s}}{2(1.25)\left(1.67×{10}^{-27}\mathrm{kg}\right)\left(3×{10}^8\mathrm{m}/\mathrm{s}\right)} \\ =\frac{6.626×{10}^{-34}\mathrm{J}·\mathrm{s}}{2.5\left(5.01×{10}^{-19}\mathrm{kg}·\mathrm{m}/\mathrm{s}\right)} \\ =\frac{6.626×{10}^{-34}\mathrm{J}·\mathrm{s}}{\left(1.2525×{10}^{-18}\mathrm{kg}·\mathrm{m}/\mathrm{s}\right)} \\ =5.29×{10}^{-16}\mathrm{J}·{\mathrm{s}}^2/\mathrm{kg}·\mathrm{m} \end{gathered}$$

Hence, further as.

$$\begin{gathered} \mathrm{λ}=5.29×{10}^{-16}\mathrm{J}·{\mathrm{s}}^2/\mathrm{kg}·\mathrm{m}×\frac{\mathrm{kg}·{\mathrm{m}}^2/{\mathrm{s}}^2}{1\mathrm{J}} \\ =5.29×{10}^{-16}\mathrm{kg}·{\mathrm{m}}^2{\mathrm{s}}^2/\mathrm{kg}·\mathrm{m}·{\mathrm{s}}^2 \\ =5.29×{10}^{-16}\mathrm{m} \end{gathered}$$

Thus, the wavelength of the photon if the pair is stationary after creation is $=5.29×{10}^{-16}$

The equation of the velocity of the photon and the lead plate is expressed as:

$$\begin{gathered} \frac{h}{\mathrm{λ}}=m_{pb}{u}_{pb} \\ {u}_{pb}=\frac{h}{m_{pb}\mathrm{λ}} \end{gathered}$$

Here, $u_{pb}$is the velocity of the photon and the lead plate, $\mathrm{λ}$is the wavelength in the first case, localid="1657593020754" $m_{pb}$is the Planck’s constant and is the mass of the combined proton and the lead nucleus.

Substitute localid="1657593961895" $6.626×{10}^{-34}\mathrm{kg}·{\mathrm{m}}^2/\mathrm{s}\text{for}h,207.2\mathrm{u}×1.67×{10}^{-27}\mathrm{kg}/\mathrm{u}\text{for}{m}_{pb}$and$6.61×{10}^{-16}\mathrm{m}$for $\mathrm{λ}$in the above equation.

$$\begin{gathered} u_{pb}=\frac{6.626×{10}^{-34}\mathrm{kg}·{\mathrm{m}}^2/\mathrm{s}}{\left(207.2\mathrm{u}×1.67×{10}^{-27}\mathrm{kg}/\mathrm{u}\right)\left(6.61×{10}^{-16}\mathrm{m}\right)} \\ =\frac{6.626×{10}^{-34}\mathrm{kg}·{\mathrm{m}}^2/\mathrm{s}}{\left(2.28×{10}^{-40}\mathrm{kg}·\mathrm{m}\right)} \\ =2.9×{10}^6 \end{gathered}$$

The equation of the energy of the proton is expressed as:

$E_p=\frac{hc}{\mathrm{λ}}$

Here, $E_p$is the energy of the proton, $h$is the Planck’s constant and $c$is the velocity of the light.

Substitute the values in the above equation.

$$\begin{gathered} E_p=\frac{\left(6.626×{10}^{-34}\mathrm{kg}·{\mathrm{m}}^2/\mathrm{s}\right)\left(3×{10}^8\mathrm{m}/\mathrm{s}\right)}{6.61×{10}^{-16}\mathrm{m}} \\ =\frac{\left(1.98×{10}^{-25}\mathrm{kg}·{\mathrm{m}}^3/{\mathrm{s}}^2\right)}{6.61×{10}^{-16}\mathrm{m}} \\ =3.007×{10}^{-10}\mathrm{kg}·{\mathrm{m}}^2/{\mathrm{s}}^2 \end{gathered}$$

The equation of the fraction of the photon’s energy in the first case is expressed as:

$E=\frac{\frac{1}{2}{m}_{pb}{u}_{pb}^2}{E_p}âˆ\pounds$

Here, $E_p$is the energy of the proton, $u_{pb}$is the velocity of the photon and the lead plate and$m_{pb}$is the mass of the combined proton and the lead nucleus.

Substitute the values in the above equation.

$$\begin{gathered} E=\frac{\frac{1}{2}\left(207.2\mathrm{u}×1.67×{10}^{-27}\mathrm{kg}/\mathrm{u}\right){\left(2.9×{10}^6\mathrm{m}/\mathrm{s}\right)}^2}{3.007×{10}^{-10}\mathrm{kg}·{\mathrm{m}}^2/{\mathrm{s}}^2} \\ =\frac{\frac{1}{2}\left(3.46×{10}^{-25}\mathrm{kg}\right)\left(8.41×{10}^{12}{\mathrm{m}}^2/{\mathrm{s}}^2\right)}{3.007×{10}^{-10}\mathrm{kg}·{\mathrm{m}}^2/{\mathrm{s}}^2} \\ =\frac{\left(1.45×{10}^{-12}\mathrm{kg}·{\mathrm{m}}^2/{\mathrm{s}}^2\right)}{3.007×{10}^{-10}\mathrm{kg}·{\mathrm{m}}^2/{\mathrm{s}}^2} \\ =0.00483 \end{gathered}$$

The equation of the velocity of the photon and the lead plate is expressed as:

$$\begin{gathered} \frac{h}{\mathrm{λ}}=m_{pb}{u}_{pb} \\ {u}_{pb}=\frac{h}{m_{pb}\mathrm{λ}} \end{gathered}$$

Here, $u_{pb}$is the velocity of the photon and the lead plate, $\mathrm{λ}$is the wavelength in the second case, $h$is the Planck’s constant and localid="1657594958229" $m_{pb}$is the mass of the combined proton and the lead nucleus.

Substitute $6.626×{10}^{-34}\mathrm{kg}·{\mathrm{m}}^2/\mathrm{s}\text{for}h,207.2\mathrm{u}×1.67×{10}^{-27}\mathrm{kg}/\mathrm{u}\text{for}{m}_{pb}$and

$5.28×{10}^{-16} m$for $\mathrm{λ}$in the above equation.

$$\begin{gathered} u_{pb}=\frac{6.626×{10}^{-34}\mathrm{kg}·{\mathrm{m}}^2/\mathrm{s}}{\left(207.2\mathrm{u}×1.67×{10}^{-27}\mathrm{kg}/\mathrm{u}\right)\left(5.28×{10}^{-16}\mathrm{m}\right)} \\ =\frac{6.626×{10}^{-34}\mathrm{kg}·{\mathrm{m}}^2/\mathrm{s}}{\left(1.82×{10}^{-40}\mathrm{kg}·\mathrm{m}\right)} \\ =3.7×{10}^{6}m/s \end{gathered}$$

The equation of the energy of the proton is expressed as:

$E_p=\frac{hC}{\mathrm{λ}}âˆ\pounds$

Here, $E_p$is the energy of the proton, $h$is the Planck’s constant and $c$is the velocity of the light.

Substitute the values in the above equation.

$$\begin{gathered} E_p=\frac{\left(6.626×{10}^{-34}\mathrm{kg}·{\mathrm{m}}^2/\mathrm{s}\right)\left(3×{10}^8\mathrm{m}/\mathrm{s}\right)}{5.28×{10}^{-18}\mathrm{m}} \\ =\frac{\left(1.98×{10}^{-25}\mathrm{kg}·{\mathrm{m}}^3/{\mathrm{s}}^2\right)}{5.28×{10}^{-18}\mathrm{m}} \\ =3.75×{10}^{-10}\mathrm{kg}·{\mathrm{m}}^2/{\mathrm{s}}^2 \end{gathered}$$

The equation of the fraction of the photon’s energy in the second case is expressed as:

$E=\frac{\frac{1}{2}{m}_{pb}{u}_{pb}^2}{E_p}$

Here, $E_p$is the energy of the proton,$u_{pb}$is the velocity of the photon and the lead plate and $m_{pb}$is the mass of the combined proton and the lead nucleus.

Substitute the values in the above equation.

$$\begin{gathered} E=\frac{\frac{1}{2}\left(207.2\mathrm{u}×1.67×{10}^{-27}\mathrm{kg}/\mathrm{u}\right){\left(3.7×{10}^6\mathrm{m}/\mathrm{s}\right)}^2}{3.75×{10}^{-10}\mathrm{kg}·{\mathrm{m}}^2/{\mathrm{s}}^2} \\ =\frac{\frac{1}{2}\left(3.46×{10}^{-25}\mathrm{kg}\right)\left(1.369×{10}^{13}{\mathrm{m}}^2/{\mathrm{s}}^2\right)}{3.75×{10}^{-10}\mathrm{kg}·{\mathrm{m}}^2/{\mathrm{s}}^2} \\ =\frac{\left(2.36×{10}^{-12}\mathrm{kg}·{\mathrm{m}}^2/{\mathrm{s}}^2\right)}{3.75×{10}^{-10}\mathrm{kg}·{\mathrm{m}}^2/{\mathrm{s}}^2} \\ =0.00627 \end{gathered}$$

Thus, the fraction of the photon energy must be absorbed by a lead nucleus in each case are$0.00483$ and$0.00627$ respectively.