Question

Calculate the polarisability and relative permittivity in Hydrogen gas with a density of 9.8 × 10 26 atoms / m3. Given the radius of the hydrogen atom is 0.50 ×10-10 m, є0 =8.85×10-12.
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Answer 1

Answer:

Explanation:

Answer Attached Below

Answer 2

Step-by-step Explanation:

Given: Number density of Atoms (N) = 9.8 × 10²⁶ atoms/m³

The radius of the Hydrogen atom = 0.50 × 10⁻¹⁰ m

To Find: Polarisability $\alpha_{e}$ and relative permittivity $\varepsilon_{r}$

Solution:

• Calculating polarisability $\alpha_{e}$

Polarisability can be calculated by the following formula;

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

Substituting given values, we will get;

$\alpha_{e} = 4 \times 3.14 \times 8.85 \times 10^{-12} \times (0.5 \times 10^{-10} )^{3} = 1.389 \times 10^{-41} \ Fm^{2}$

• Calculating relative permittivity $\varepsilon_{r}$

Polarisability $\alpha_{e}$ and relative permittivity  $\varepsilon_{r}$ are related as

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

$\Rightarrow \varepsilon_{r} = \frac{N \alpha_{e}}{\varepsilon_{0}} + 1$

Substituting the given values, we will get

$\varepsilon_{r} = \frac{9.8 \times 10^{26} \times 1.389 \times 10^{-41}}{8.85 \times 10^{-12}} + 1 = 1.0051 \ F/m$

Hence, Polarisability is $\alpha_{e} = 1.389 \times 10^{-41} \ Fm^{2}$  and relative permittivity  is $\varepsilon_{r} = 1.0051 \ F/m$