A gas molecule of mass M at the surface of the earth has kinetic energy equivalent to 0 °C. If it were to go up straight without colliding with any other molecules, how high it would rise? Assume that the height attained is much less than the radius of the earth. (k$$_B$$ is Boltzmann constant)

The gas molecule has a kinetic energy equivalent to 0 °C. Since temperature in kinetic theory is given in Kelvin, convert 0 °C to Kelvin: 0 °C = 273 K. The average translational kinetic energy of a gas molecule is given by $$\frac{3}{2} k_B T$$, where $$k_B$$ is the Boltzmann constant and $$T$$ is the absolute temperature. Therefore, the kinetic energy $$K$$ at the surface is:

$$ K = \frac{3}{2} k_B \times 273 $$

As the molecule rises straight up without colliding with other molecules, it will lose kinetic energy and gain potential energy. At the maximum height, its kinetic energy becomes zero, and all the initial kinetic energy is converted to potential energy. Assuming the potential energy at the Earth's surface is zero, by conservation of energy:

Initial kinetic energy + Initial potential energy = Final kinetic energy + Final potential energy

$$ K + 0 = 0 + U_{\text{max}} $$

So,

$$ U_{\text{max}} = K = \frac{3}{2} k_B \times 273 $$

The potential energy at height $$h$$ above the Earth's surface is $$M g h$$, where $$M$$ is the mass of the molecule, $$g$$ is the acceleration due to gravity, and $$h$$ is much less than the Earth's radius (as given). Therefore:

$$ U_{\text{max}} = M g h $$

Equating the expressions for potential energy:

$$ M g h = \frac{3}{2} k_B \times 273 $$

Solving for $$h$$:

$$ h = \frac{\frac{3}{2} k_B \times 273}{M g} $$

Simplify the fraction:

$$ h = \frac{3}{2} \times \frac{273 k_B}{M g} $$

Multiply the constants:

$$ h = \frac{3 \times 273 k_B}{2 M g} $$

Calculate $$3 \times 273$$:

$$ 3 \times 273 = 819 $$

So,

$$ h = \frac{819 k_B}{2 M g} $$

Comparing with the given options, this matches option B. Hence, the correct answer is Option B.