Dropping one at a time

Before the helium atom fell, it had a mechanical energy equal to its potential energy, where $E = mgh$ .

When the helium atom fell into the tank, its potential energy had transformed into kinetic energy, where $E = (1/2)mv^2$ .

As the mechanical energy is conserved throughout the movement (as it fell in a vacuum), the initial potential energy equals the kinetic energy when it entered the tank. $mgh = (1/2)mv^2, h = (v^2)/2g$ .

As the temperature of the helium atoms in the tank remained at 300K, the speed of the falling helium atom must be equal to the root mean square speed of the other helium atoms previously in the tank. The root mean square speed can be calculated by the formula: $v_{rms} = \sqrt{(3RT)/M} = \sqrt {(3(8.314)300)/4} = 43.251$

Insert the v(rms) into the equation for h mentioned previously: $h = (v^2)/2g = (43.251^2)/2(9.8)$ = 95.441 meters

Let the helium atom falls from a height ( h ) metres above the tank of helium atoms . Since it falls from rest its initial Kinetic energy is zero.

At this point it has energy only in the form of potential energy, U =mgh

here, m is the mass of helium atom, g is acceleration due to gravity and h is height .

According to law of conservation of energy , when the helium atom falls into the tank of gaseous helium atoms ,then all its initial potential energy (U) changes into the kinetic energy .

But it is given that the temperature of the tank still remains same. thus, kinetic energy = (3/2)(RT) where R is universal gas constant and T is temperature (i.e. 300 K) equating U = kinetic energy we will obtain , h =95.44 metres

The average kinetic energy of the helium atoms is given by $K.E.=\frac{3}{2}k_BT$ where $k_B$ is Boltzmann's constant. In order for the temperature to not change, the kinetic energy of the falling atom should be the same. Therefore $mgh=\frac{3}{2}k_B T$ . m for helium is $6.64\times 10^{-27}~kg$ and so we have h=95000.5 m.