Question

Calculate the electronic polarizability of argon atom. Given that εr=1.0024 at NTP and N=2.7x10²⁵ atoms/m³.

Answer 1

Solution :

 

Relative permitivity = 1.0024 at NTP 

Permitivity of free space  = 8.854 × 10–12 Fm–1

Answer 2

Answer:

The argon atoms's electronic polarizability, $\alpha _{e}$, calculated is  $7.86\times 10^{-40}F-m^{2}$.

Explanation:

Given,

The argon atoms's relative permeability, $\epsilon _{r}$ = $1.0024$

The total number of given atoms per unit volume, N = $2.7\times 10^{25} atoms/m^{3}$

The electronic polarizability of an argon atom, $\alpha _{e}$ =?

As we know,

• Electronic polarizability, $\alpha _{e}$ = $\frac{\epsilon _{0} ( \epsilon _{r}-1) }{N}$

Here, $\epsilon _{0}$ is permeability constant = $8.85\times 10^{-12}F/m$.

After putting the value in the equation, we get:

• $\alpha _{e}$ = $\frac{\epsilon _{0} ( \epsilon _{r}-1) }{N}$
• $\alpha _{e}$ = $\frac{8.85\times 10^{-12} ( 1.0024-1) }{2.7\times 10^{25} }$
• $\alpha _{e}$ = $\frac{8.85\times 10^{-12} ( 0.0024) }{2.7\times 10^{25} }$
• $\alpha _{e}$ = $7.86\times 10^{-40}F-m^{2}$

Hence, the electronic polarizability of an argon atom, $\alpha _{e}$ =  $7.86\times 10^{-40}F-m^{2}$.