Ncert Exercises & Solutions

Question 11.35: Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature (27 2C) and 1 atm pressure, and compare it with the mean separation between two atoms under these conditions.

Hint: $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$
Step 1: Find the average energy of gas at T.

The average energy of gas at T is given as:

E = \frac{3}{2} kT

Step 2: Find the de Broglie wavelength of the He atom.
De Broglie wavelength is given by the relation:

λ = \frac{h}{\sqrt{2 mE}}

Where, m = Mass of a He atom

\begin{aligned} \begin{matrix} = \frac{Atomic weight}{N_{A}} \\ = \frac{4}{6 .023 \times 10^{23}} \\ = 6 .64 \times 10^{- 24} g = 6 .64 \times 10^{- 27} kg \\ ∴ λ = \frac{h}{\sqrt{3 mkT}} \\ = \frac{6 .6 \times 10^{- 34}}{\sqrt{3 \times 6 .64 \times 10^{- 27} \times 1 .38 \times 10^{- 23} \times 300}} \\ = 0 .7268 \times 10^{- 10} m \end{matrix} \end{aligned}

Step 3: Find the mean distance between two atoms.
$\begin{aligned} We have the ideal gas formula: \\ PV = RT \\ \begin{array}{r} PV = kNT \\ \frac{V}{N} = \frac{kT}{P} \end{array} \end{aligned}$

The mean separation between two atoms of the gas is given by the relation:

\begin{aligned} r & = {(\frac{V}{N})}^{\frac{1}{3}} = {(\frac{kT}{P})}^{\frac{1}{3}} \\ = {[\frac{1 .38 \times 10^{- 23} \times 300}{1 .01 \times 10^{5}}]}^{\frac{1}{3}} \\ = 3 .35 \times 10^{- 9} m \end{aligned}

Hence, the mean separation between the atoms is much greater than the de Broglie wavelength.

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions