Question

The total energy of a proton passing through a laboratory apparatus is 10.611-nJ. What is its speed parameter $\mathbf{\beta }$ ? Use the proton mass given in Appendix B under “Best Value,” not the commonly remembered rounded number.

Short answer

The speed parameter is $\mathbf{0}\mathbf{.}\mathbf{9898856096}\mathbf{c}$

Step-by-step solution

Relativistic Total energy

The total energy of an object is the sum of its rest mass energy, kinetic energy if its moving, and potential energy if it is under some force.The total relativistic energy of an object moving at constant velocity will be expressed as

$E=\gamma m{c}^2$

Where, $\gamma$is the Lorentz factor, and m, is the rest mass of the object. Here in the question, it is asked to use the best value for the proton’s mass$1.672621637\times {10}^{-27}kg$.

Inserting these values in the above equation to get$\gamma$,

$$\begin{gathered} 10.611\times {10}^{-9}J=\gamma \left(1.672621637\times {10}^{-27}kg\right){\left(3\times {10}^{8}m/s\right)}^2 \\ 10.611\times {10}^{-9}J=\gamma \left(\text{1.505359473}\times {\text{10}}^{-11}kg{m}^2/s^2\right) \\ \gamma =\frac{10.611\times {10}^{-9}}{\text{1.505359473}\times {\text{10}}^{-11}} \\ \approx \text{7.048814712} \end{gathered}$$

Lorentz factor

The result of 2^nd postulate of the special theory of relativity is that the clocks run slower for a moving object when measured from a rest frame. The factor by which the clock is running differently is called the Lorentz factor.

The expression for Lorentz factor is

$\gamma =\frac{1}{\sqrt{1-{\beta }^2}}$

Here $\beta$is the speed parameter$v/c$. Inserting the value of the Lorentz factor in this expression

$$\begin{gathered} \gamma =\frac{1}{\sqrt{1-{\beta }^2}} \\ \beta =\sqrt{1-\frac{1}{{\gamma }^2}} \\ =\sqrt{1-\frac{1}{{\left(7.048814712\right)}^2}} \\ =\text{0.989885609}6 \end{gathered}$$

The speed parameter thus obtained up to 10 significant values because the best value of proton’s mass is given in 10 significant values is asked to be considered.