What is the de broglie wavelength of He at 25 celcius?

We know that the de broglie wavelength is given by

l = h/p = h/mv

We know that the mass of 1 mole of He atoms is 4 g. 

Therefore, mass of one He atom = 4/(6.023 x 10²³) = 6.64 x 10^-24 g

Using the Kinetic theory of gases, we can relate temperature to velocity of molecules.

The Kinetic theory of gases says that in thermal equilibrium, the Kinetic energy of one mole of atoms in a gas is given by

KE = (3/2)RT

Therefore, KE of one mole of He atoms = (3/2)* 8.314*(25+273) [Convert 25 C to K]

= 3716.358 J

Hence, Kinetic Energy of one He atom = 3716.358/(6.023 x 10²³) = 6.17 x 10^-21 J

Now, the Kinetic Energy = (1/2)mv² .

So, 6.17 x 10^-21 = (1/2)(6.64 x 10^-24)v²

v² = 1858.433

So, v = sqrt(1858.433) = 43.11 m/s

Hence momentum mv = 6.64 x 10^-24 * 43.11 = 2.86 x 10^-22 kg.m/s

So, finally, de broglie wavelength l = h/p = 6.626 x 10^-34 / 2.86 x 10^-22 = 2.31 x 10^-12 m