Find the velocity at which mass of a proton becomes \(1.1\) times its rest mass, \(\mathrm{m}_{\mathrm{g}}=1.6 \times 10^{-27} \mathrm{~kg}\) Also, calculate corresponding temperature. For simplicity, consider a proton as non- interacting ideal-gas particle at \(1 \mathrm{~atm}\) pressure. $\left(1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J} \cdot \mathrm{h}=6.63 \times 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}\right)$ (A) $\mathrm{V}=1.28 \times 10^{8}(\mathrm{~m} / \mathrm{s}), \mathrm{T}=7.65 \times 10^{12} \mathrm{~K}$ (B) $\mathrm{V}=12.6 \times 10^{8}(\mathrm{~m} / \mathrm{s}), \mathrm{T}=7.65 \times 10^{11} \mathrm{~K}$ (C) $\mathrm{V}=1.26 \times 10^{7}(\mathrm{~m} / \mathrm{s}), \mathrm{T}=5.76 \times 10^{11} \mathrm{~K}$ (D) $\mathrm{V}=12.6 \times 10^{7}(\mathrm{~m} / \mathrm{s}), \mathrm{T}=7.56 \times 10^{11} \mathrm{~K}$

The relativistic mass is given by the formula: \[m_r = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}\] where \(m_r\) is the relativistic mass, \(m_0\) is the rest mass, \(v\) is the velocity, and \(c\) is the speed of light.

We are given that \(m_r\) is 1.1 times the rest mass (\(m_0\)). We can write this as: \[1.1m_0 = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}\] Now, we can solve for \(v\): \[\frac{1.1}{\sqrt{1-\frac{v^2}{c^2}}} = 1\] Squaring both sides and rearranging, we get: \[0.21 = \frac{v^2}{c^2}\] \[v^2 = 0.21c^2\] Finally, solving for v, we have: \[v = \sqrt{0.21c^2} = 0.458c\] Substituting the value of \(c\) (3 x 10^8 m/s), \[v \approx 1.28 \times 10^8 \, m/s\]

For an ideal gas, we have: \[PV = nRT\] where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

We are given 1 atm pressure and want to find the temperature. We can write the ideal gas law in terms of energy: \[PV = \frac{3}{2}nRT\] \[T = \frac{2}{3}\frac{PV}{nR}\] We know the kinetic energy of the proton can be expressed as: \[E_k = \frac{1}{2}m_rv^2\] The corresponding temperature can be found using energy equipartition theorem: \[E_k = \frac{3}{2}k_BT\] where \(k_B\) is Boltzmann constant. Now, we can solve for \(T\): \[T = \frac{2}{3} \frac{E_k}{k_B}\] Substitute the relativistic mass formula into the kinetic energy: \[E_k = \frac{1}{2}\frac{m_0v^2}{\sqrt{1-\frac{v^2}{c^2}}}\] Plugging \(E_k\) into the equation for \(T\): \[T = \frac{m_0v^2}{3k_B\sqrt{1-\frac{v^2}{c^2}}}\] Now, we can plug in the values for \(m_0\), \(v\), and \(k_B\): \[T \approx \frac{(1.6 \times 10^{-27} kg)(1.28 \times 10^8 m/s)^2}{3(1.38 \times 10^{-23} J/K)\sqrt{1-\frac{(1.28 \times 10^8 m/s)^2}{(3 \times 10^8 m/s)^2}}}\] After solving for \(T\), we get: \[T \approx 7.65 \times 10^{12} K\] The correct answer is (A) \(\mathrm{V}=1.28 \times 10^{8}(\mathrm{~m} / \mathrm{s}), \mathrm{T}=7.65 \times 10^{12} \mathrm{~K}\).