Question

The relative permittivity of Argon at 0C and one atmosphere is 1.000435. Calculate the polarizability of the atom

Answer 1

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Answer 2

Answer:

$1.43\times10^{-40} Fm^{2}$

Explanation:

Given:

$\epsilon_r=1.0004355$

To find:

Polarizability

Solution:

The characteristic of a substance or a medium known as permittivity has an impact on the coulomb force between two point charges when that substance or medium is used. In a vacuum, coulomb force is greatest. You can measure the other medium's force in relation to it. Relative permittivity of the medium is the quantity that describes the amount by which the coulombic force between two charges is reduced in comparison to vacuum.

The ratio of the medium's absolute permittivity to the vacuum's permittivity can also be used to determine relative permittivity. The term "dielectric constant" also refers to relative permittivity.

The response of a molecule to an external field is determined by polarizability.

Atoms of argon at NTP in number = $\frac{6.023\times10^{26} }{22.4}$

=$2.69\times10^{25}$

From the formula,

$\epsilon_o(\epsilon_r-1)=N\alpha _e$

Substituting the values

$\alpha _e=\frac{\epsilon_o(\epsilon_r-1)}{N} \\=\frac{8.85\times10^{-12}\times1.000435 }{2.69\times10^{25} }$

=$1.43\times10^{-40} Fm^{2}$

What is permittivity and relative permittivity? How are they related?

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Define the term polarizability​

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