• We know that the electronic polarizability$\mathrm{a_e}$ is

$\mathrm{a_e = \dfrac{\varepsilon_0(\varepsilon_r-1)}{N}}$

• Substituting the values, we get

$\mathrm{=\dfrac{8.85 \times 10^{-12} \times (1.0024 -1)}{2.7 \times 10^{25}}}$

$\mathrm{= 7.866 \times 10^{-40} F m^2}$

• Therefore, the electonic polarisability of air atom at NTP is equal to$\mathrm{7.866 \times 10^{-40} F m^2}$