JEE Challenger
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PhyMedAdv 2011 • Paper 1

Estimate Plasma Reflection Wavelength for Given Electron Density

Comprehension Passage

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 begin to oscillate about the positive ions with a natural angular frequency ‘ωp\omega_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\omega, where a part of the energy is absorbed and a part of it is reflected. As ω\omega approaches ωp\omega_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.

Estimate the wavelength at which plasma reflection will occur for a metal having the density of electrons N≈4×1027 m−3N \approx 4 \times 10^{27}\text{ m}^{-3}. Take ε0≈10−11\varepsilon_0 \approx 10^{-11} and m≈10−30m \approx 10^{-30}, where these quantities are in proper SI units.

Options

A

800 nm800\text{ nm}

B

600 nm600\text{ nm}

Correct
C

300 nm300\text{ nm}

D

200 nm200\text{ nm}

Topics & Concepts

Electromagnetic WavesElectromagnetic Waves

Step-by-Step Solution

The natural plasma frequency ωp\omega_p is given by the expression derived from dimensional analysis and plasma physics:
ωp=Ne2mε0\omega_p = \sqrt{\frac{N e^2}{m \varepsilon_0}}

Given the values in SI units:

  • Electron number density, N≈4×1027 m−3N \approx 4 \times 10^{27} \text{ m}^{-3}
  • Charge of an electron, e≈1.6×10−19 Ce \approx 1.6 \times 10^{-19} \text{ C}
  • Permittivity of free space, ε0≈10−11 F/m\varepsilon_0 \approx 10^{-11} \text{ F/m}
  • Mass of an electron, m≈10−30 kgm \approx 10^{-30} \text{ kg}
  • Speed of light, c≈3×108 m/sc \approx 3 \times 10^8 \text{ m/s}

Substituting these values into the formula for ωp\omega_p:
ωp=(4×1027)×(1.6×10−19)210−30×10−11\omega_p = \sqrt{\frac{(4 \times 10^{27}) \times (1.6 \times 10^{-19})^2}{10^{-30} \times 10^{-11}}}

ωp=4×1027×2.56×10−3810−41\omega_p = \sqrt{\frac{4 \times 10^{27} \times 2.56 \times 10^{-38}}{10^{-41}}}

ωp=10.24×1030≈3.2×1015 rad/s\omega_p = \sqrt{10.24 \times 10^{30}} \approx 3.2 \times 10^{15} \text{ rad/s}

Plasma reflection occurs at resonance when the incident electromagnetic wave frequency ω\omega equals the plasma frequency ωp\omega_p. The corresponding wavelength λ\lambda is given by:
λ=2πcωp\lambda = \frac{2\pi c}{\omega_p}

Substituting the known parameters:
λ=2π×(3×108)3.2×1015\lambda = \frac{2 \pi \times (3 \times 10^8)}{3.2 \times 10^{15}}

λ≈6×3.14×1083.2×1015=5.8875×10−7 m≈600 nm\lambda \approx \frac{6 \times 3.14 \times 10^8}{3.2 \times 10^{15}} = 5.8875 \times 10^{-7} \text{ m} \approx 600 \text{ nm}

Thus, the estimated wavelength for plasma reflection is approximately 600 nm600\text{ nm}.

Correct Option: B

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