91Ó°ÊÓ

The total energy of proton is E = 14.24 nj

The left right width in measurement is $L_0=21\text{cm}=0.21\text{m}$

The Lorentz factor is calculated by using the formula for total relativistic energy of proton then length contraction and time dilation formulas are used to find according to different frames.

(a)

The Lorentz factor for proton is calculated as:

$E=\mathrm{γ}m{c}^2$

Here, m is the mass of electron and its value is $1.67×{10}^{-27}\text{kg}$, c is the speed of light and its value is $3×{10}^8\text{m}/\text{s}$

$$\begin{gathered} \left(14.24\text{nJ}\right)\left(\frac{{10}^{-9}\text{J}}{1\text{nJ}}\right)=\mathrm{γ}\left(1.67×{10}^{-27}\text{kg}\right){\left(3×{10}^8\text{m}/\text{s}\right)}^2 \\ \mathrm{γ}=94.74 \end{gathered}$$

The width in proton’s frame is given as:

$L=\frac{L_0}{\mathrm{γ}}$

Substitute all the values in equation.

$$\begin{gathered} L=\frac{21\text{cm}}{94.74} \\ L=0.222\text{cm} \end{gathered}$$

Therefore, the width in proton’s frame is 0.222 m.

(b)

The time of passage for proton in rest frame is given as:

$t_r=\frac{L_0}{\left(\sqrt{1-\frac{1}{{\mathrm{γ}}^2}}\right)c}$

Substitute all the values in above equation.

$$\begin{gathered} \frac{21\text{cm}}{\left(\sqrt{1-\frac{1}{{\left(94.74\right)}^2}}\right)\left(3×{10}^{10}\text{cm}/\text{s}\right)} \\ {t}_r=7.01×{10}^{-10}\text{s}l \\ {t}_r=\left(7.01×{10}^{-10}\text{s}\right)\left(\frac{1\text{ps}}{{10}^{-12}\text{s}}\right) \\ {t}_r=701\text{ps} \end{gathered}$$

Therefore, the time of passage for proton in rest frame is 701 ps .

(c)

The time of passage for proton in proton’s frame is given as:

$t=\frac{t_r}{\mathrm{γ}}$

Substitute all the values in above equation.

$$\begin{gathered} t=\frac{701\text{ps}}{94.74} \\ t=7.40\text{ps} \end{gathered}$$

Therefore, the time of passage for proton in proton’s frame is 7.40ps .