Question

Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature $\left( {27^\circ C} \right)$and 1atm pressure, and compare it with the mean separation between two atoms under these conditions.

1. ${r_o} = 6.4 \times {10^{ - 9}}m$
2. ${r_o} = 3.4 \times {10^{ - 9}}m$
3. ${r_o} = 4.4 \times {10^{ - 9}}m$
4. ${r_o} = 5.4 \times {10^{ - 9}}m$

Answer 1

In the question we have asked to compare the De-Broglie wavelength which is associated with the He atom with the mean separation between two atoms. De – Broglie wavelength is the starting of quantum mechanics. It determines the probability density of finding the object at a given time in a specified space.

Step by step solution:
Step 1: Calculate the De-Broglie wavelength
Find out the mass of the helium atom:
$m = \dfrac{{{\text{Atomic mass of He}}}}{{{\text{Avogadro's Number}}}}$ ;
Put in the given value in the above equation:
$\Rightarrow m = \dfrac{4}{{6 \times {{10}^{23}}}}gms$;
$\Rightarrow m = 6.67 \times {10^{ - 27}}kg$;
De-Broglie wavelength is given as:
$\lambda = \dfrac{h}{p}$ ;
Where:
$\lambda$= Wavelength;
$h$= Planck’s Constant;
$p$= momentum;
$\Rightarrow \lambda = \dfrac{h}{{\sqrt {3mkT} }}$;
Here:
$p = \sqrt {3mkT}$ ;
m = Mass;
k = Constant;
T= Absolute Temperature;

Put in the given value in the above equation:
$\Rightarrow \lambda = \dfrac{{6.63 \times {{10}^{ - 34}}}}{{\sqrt {3 \times 6.67 \times {{10}^{ - 27}} \times 1.38 \times {{10}^{ - 23}} \times 300} }}$;
$\Rightarrow \lambda = 7 \times {10^{ - 11}}m$;

Step 2: Calculate the mean separation
The kinetic gas equation for 1mole of gas is given as:
$PV = RT = kNT$;
Where:
P = Pressure;
V = Volume;
T = Absolute Temperature;
R = Ideal gas constant;
N = Number of atoms and molecules;
k = Constant;

$\Rightarrow \dfrac{V}{N} = \dfrac{{kT}}{P}$;
The mean separation is given by:
${r_o} = {\left( {\dfrac{{{\text{Molar Volume}}}}{{{\text{Avogadro's Number}}}}} \right)^{\dfrac{1}{3}}}$ ;
${r_o} = {\left( {\dfrac{{\text{V}}}{N}} \right)^{\dfrac{1}{3}}}$;
Now put $\dfrac{V}{N} = \dfrac{{kT}}{P}$in the above equation and solve:
$\Rightarrow {r_o} = {\left( {\dfrac{{{\text{KT}}}}{P}} \right)^{\dfrac{1}{3}}}$;
Put in the given values in the above equation:
$\Rightarrow {r_o} = {\left( {\dfrac{{{\text{1}}{\text{.38}} \times {\text{1}}{{\text{0}}^{ - 23}} \times 300}}{{1.01 \times {{10}^5}}}} \right)^{\dfrac{1}{3}}}$;
The mean separation is:
$\Rightarrow {r_o} = 3.4 \times {10^{ - 9}}m$;

Option “2” is correct. The mean separation is ${r_o} = 3.4 \times {10^{ - 9}}m$. Here the mean separation is very large as compared to the de Broglie wavelength.

Note: Here, we have to first find out the De-Broglie wavelength by first finding out the mass of the helium atom and then put the mass in the famous equation of the De-Broglie wavelength. Then apply the formula for mean separation and make a relation with the formula for gas constant PV = RT = kNT.