Question

A dense collection of equal number of electrons and positive ions is called neutral plasma. Certain solids containing fixed positive ions surrounded by free electrons can be treated as neutral plasma. Let 'N' be the number density of free electrons, each of mass 'm'. When the electrons are subjected to an electric field, they are displaced relatively away from the heavy positive ions. If the electric field becomes zero, the electrons being to oscillate about the positive ions with a natural angular frequency ' $ω$ _p', which is called the plasma frequency. To sustain the oscillations, a time varying electric field needs to be applied that has an angular frequency $ω$ , where a part of the energy is absorbed and a part of it is reflected. As $ω$ approaches $ω$ _p, all the free electrons are set to resonance together and all the energy is reflected. This is the explanation of high reflectivity of metals.

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Linked Question 1

Taking the electronic charge as ' $ε$ ' and the permittivity as ' $ε$ ₀' , use dimensional analysis to determine the correct expression for $ω$ _p.

$\sqrt{\frac{{mε}_{0}}{Ne}}$

$\sqrt{\frac{{Ne}^{2}}{{mε}_{0}}}$

$\sqrt{\frac{{mε}_{0}}{{Ne}^{2}}}$

$\sqrt{\frac{Ne}{{mε}_{0}}}$

${F=mω}^{2} ℓ \equiv \frac{e^{2}}{4 {πε}_{0} ℓ^{2}}$

$ω^{2} \equiv \frac{e^{2}}{4 {πε}_{0} ℓ^{3}} \equiv (\frac{e}{{mε}_{0}}) (\frac{{Nℓ}^{3}}{ℓ^{3}})$

$ω = \sqrt{\frac{{Ne}^{2}}{{mε}_{0}}}$

Linked Question 2

Estimate the wavelength at which plasma reflection will occur for a metal having the density of electrons N $\approx$ 4 × 10²⁷ m^–3. Take e₀ $\approx$ 10^–11 and m $\approx$ 10^–30, where these quantities are in proper SI units.

300 nm

800 nm

600 nm

200 nm

$c = λf$

$ω_{p} = ω = \frac{2 πc}{λ} = \sqrt{\frac{{Ne}^{2}}{{mε}_{0}}}$

$λ = 2 πc \sqrt{\frac{{mε}_{0}}{{Ne}^{2}}} = \frac{2 πc}{e} \sqrt{\frac{{mε}_{0}}{N}} = \frac{2 \times 3 .14 \times 10^{8}}{1 .6 \times 10^{- 19}} \sqrt{\frac{(10^{- 30}) (10^{- 11})}{4 \times 10^{27}}}$

= 589 × 10^–9 m $\approx$ 600 nm