91Ó°ÊÓ

The relativistic formula is $\frac{\mathrm{v}}{\mathrm{c}}=\sqrt{1-{\left(\frac{0.511}{\mathrm{V}}\right)}^2}$.

The series of Binomial series used in this solution is given:

$\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{x}{\mathbf{)}}^{\mathbf{1}\mathbf{/}\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathit{x}\mathbf{-}\frac{\mathbf{1}}{\mathbf{8}}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{16}}{\mathit{x}}^{\mathbf{3}}\mathbf{-}\frac{\mathbf{5}}{\mathbf{128}}{\mathit{x}}^{\mathbf{4}}\mathbf{+}\mathbf{…}$

Write the binomial series to be used.

$(1+x{)}^{1/2}=1+\frac{1}{2}x-\frac{1}{8}{x}^2+\frac{1}{16}{x}^3-\frac{5}{128}{x}^4+…$

Truncate the binomial series after two terms and use it to express the relativistic formula.

$$\begin{gathered} \frac{v}{c}=\sqrt{1-{\left(\frac{0.511}{V}\right)}^2} \\ =1-\frac{1}{2}{\left(\frac{0.511}{V}\right)}^2 \\ \frac{v}{c}=1-\frac{0.1305605}{V^2} \\ 1-\frac{v}{c}=\frac{0.1305605}{V^2} \end{gathered}$$

Substitute $V=100$million volts in the equation above.

$$\begin{gathered} 1-\frac{v}{c}=\frac{0.1305605}{{\left(100\right)}^2} \\ 1-\frac{v}{c}=1.3×{10}^{-5} \end{gathered}$$

Thus, the answer is $1.3×{10}^{-5}$.

Substitute $V=500$million volts in the equation, and we get,

$$\begin{gathered} 1-\frac{v}{c}=\frac{0.1305605}{{\left(500\right)}^2} \\ 1-\frac{v}{c}=5.22242×{10}^{-7} \end{gathered}$$

Thus, the answer is $5.22242×{10}^{-7}$.

Substitute $V=25000$million volts in the equation above.

role="math" localid="1662643138333" $$\begin{gathered} 1-\frac{v}{c}=\frac{0.1305605}{{\left(25000\right)}^2} \\ 1-\frac{v}{c}=2.08897×{10}^{-10} \end{gathered}$$

Thus, the answer is$2.08897×{10}^{-10}$.

Substitute $V=100000$million volts in the equation above.

role="math" localid="1662643365794" $\begin{array}{rcl}1-\frac{v}{c}& =& \frac{0.1305605}{{\left(100000\right)}^2}\\ 1-\frac{v}{c}& =& 1.3056×{10}^{-11}\\ & & \end{array}$

Thus, the answer is $1.3056×{10}^{-11}$.