Question

Find the velocity at which mass of a proton becomes 1.1 times its rest mass, mp = 1.6 x 10⁻²⁷ kg. Also, calculate corresponding temperature. For simplicity, consider a proton as non-interacting ideal-gas particle at 1 atm pressure.[c = 3 x 10⁸ ms⁻¹ , kB = 1.38 x 10⁻²³ SI]

Answer 1

Answer: velocity = 1.24*10^8 m/s and temperature is 4830.2 kelvin.

Explanation:

Answer 2

The velocity of proton is $\bold{1.28 \times 10^{8} \ \mathrm{m} / \mathrm{s}}$

The corresponding temperature is $\bold{7.65 \times 10^{12} \ \mathrm{K}}$

Given:

Mass of a proton becomes 1.1 times its rest mass:

$\bold{\Rightarrow m=1 \cdot 1 \ m_{0}}$

mp = 1.6 × 10⁻²⁷ kg

To find:

Velocity of proton = ?

Corresponding temperature = ?

Solution:

Velocity of the proton:

$\bold{\mathrm{m}=\frac{\mathrm{m}_{\mathrm{o}}}{ \sqrt{1-\frac{\mathrm{V}^{2}}{ \mathrm{c}^{2}}\right}}}$

On bringing mass on one side, we get,

$\bold{\sqrt{1-\frac{\mathrm{V}^{2}}{\mathrm{c}^{2}}\right)}=\frac{\mathrm{m}_{\mathrm{o}}}{\mathrm{m}}}$

From question,

$\bold{\Rightarrow m=1 \cdot 1 \ m_{0}}$

$\bold{1-\frac{V^{2}}{c^{2}}=\left(\frac{m_{o}}{1.1 \ m_{o}}\right)^{2}}$

$\bold{0.173=\mathrm{V}^{2} / \mathrm{c}^{2}}$

$\bold{\mathrm{v}^{2}=0.173 \times \mathrm{c}^{2}}$

Where, c = Velocity of light $\bold{=3 \times 10^{8} \ m/s}$

$\bold{v^2=0.173 \times\left(3 \times 10^{8}\right)^{2}}$

$\bold{\mathrm{V}=1.28 \times 10^{8} \ \mathrm{m} / \mathrm{s}}$

Temperature of the proton:

Energy is given by the formula:

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

$\bold{E=1 \cdot 1 \mathrm{m}_{\mathrm{o}} \mathrm{c}^{2}}$

On substituting the know values, we get,

$\bold{E=1 \cdot 1 \times 1 \cdot 6 \times 10^{-27} \times\left(3 \times 10^{8}\right)^{2}}$

$\bold{E=1.58 \times 10^{-10} \ \mathrm{J}}$

Energy and the temperature is related by the formula given below:

$\bold{E=\frac{3}{2} K T}$

$\bold{\mathrm{T}=\frac{2}{3}(\mathrm{E} / \mathrm{K})}$

$\bold{\Rightarrow T= \frac{2}{3}\times \frac{1.58 \times 10^{-10}}{1.38 \times 10^{-23}}}$

$\bold{\therefore T = 7.65 \times 10^{12} \ \mathrm{K}}$