​A proton with rest mass of 1.67 × $$10^{-27}$$​ kg moves in an accelerator with a speed of 0.6 c. Its total energy is ($$c$$​ = 3 × $$10^8$$​ m/s)​

Correct option is A

​$$\begin{aligned}&\text{Given:} \\&\bullet \ \text{Rest mass of proton } m_0 = 1.67 \times 10^{-27} \, \text{kg} \\&\bullet \ \text{Speed } v = 0.6c = 0.6 \times 3 \times 10^8 = 1.8 \times 10^8 \, \text{m/s} \\&\bullet \ \text{Speed of light } c = 3 \times 10^8 \, \text{m/s} \\[10pt]&E = \frac{m_0 c^2}{\sqrt{1 - \frac{v^2}{c^2}}} \\[10pt]&\text{ } \frac{v^2}{c^2} = (0.6)^2 = 0.36 \\[6pt]&\text{ So,} \\&E = \frac{1.67 \times 10^{-27} \times (3 \times 10^8)^2}{\sqrt{1 - 0.36}} = \frac{1.67 \times 9 \times 10^{-11}}{\sqrt{0.64}} \\[6pt]&E = \frac{15.03 \times 10^{-11}}{0.8} = 18.79 \times 10^{-11} \, \text{J} \\[6pt]&E = 1.879 \times 10^{-10} \, \text{J} \approx \boxed{1.88 \times 10^{-10}} \, \text{J}\end{aligned}$$​​