Calculate electronic polarizability of Argon atom. Given relative permittivity is 1.0024 at NTP and the gas contains 2.7 × 1025 atoms per m3.

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written 3.3 years ago by

teamques10 ★ 66k

• modified 3.3 years ago

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written 3.3 years ago by

teamques10 ★ 66k

modified 2.8 years ago by

sagarkolekar ★ 11k

GIVEN:

$\in_{0} = 1.0024$ at NTP

N =$ 2.7 \times 10^{25} /m^{3}$

FORMULA: $ \alpha_{e} = \dfrac{\in_{0} (\in_{r} - 1)}{N}$

CALCULATION :

$\alpha_{e} = \dfrac{\in_{0} (\in_{r} - 1)}{N}$

=$ \dfrac{8.85\times10^{-12}(1.0024 - 1)}{2.7\times10^{25}}$

= $7.8667\times10^{-40}$ F.M²

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