Question

The number of atoms in hydrogen gas is 9.8 × 1020atoms /cc. The radius of hydrogen
atom is 0.053 nm. Calculate the electronic polarizability and relative permittivity

Answer 1

Answer:

sorry but I don't know the answer

Answer 2

Answer:

The electronic polarizability and relative permittivity is 0.165*10⁻⁴⁰Fm² and ∈=1.182x10⁻⁶Fm⁻¹

Explanation:

Given:

No. of atoms in the gas $\mathrm{N}=9.8 \times 10^{20} atoms / \mathrm{cc}^{}$

Permitivity of free space $\varepsilon_{0}=8.854 \times 10^{-12} \mathrm{Fm}^{-1}$

1. Electronic polarizability

$\alpha_{e}=4 \pi \varepsilon_{0} R^{3}$

     =4XπX8.854X10⁻¹²X(.053*10⁻⁹)³

      =0.165*10⁻⁴⁰

2. Polarization

$\begin{aligned}P &=\varepsilon_{0}\left(\varepsilon_{\mathrm{r}}-1\right) \mathrm{E} \\\text { and }\\ \mathrm{P} &=N \alpha_{\mathrm{e}} \mathrm{E}\end{aligned}$

From the above two equations, we can write

$N \alpha_{e}=\varepsilon_{0}\left(\varepsilon_{\mathrm{r}}-1\right) \\\\ \alpha_{e}=\frac{\varepsilon_{0}\left(\varepsilon_{\mathrm{r}}-1\right)}{\mathrm{N}}$

0.165*10⁻⁴⁰=8.854x10⁻¹²(∈-1)/9.8x10²⁰

∈=1.182x10⁻⁶Fm⁻¹