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Magnetism question

2011 · Shift 1 · Q48
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  5. /2011 · Shift 1 · Q48

Magnetism question

2011 · Shift 1 · Q48

JEE AdvancedPhysicsMagnetismMCQ+3 / −0.75
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}ωp​' 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\omegaω, where a part of the energy is absorbed and a part of it is reflected. As ω\omegaω approaches ωp{\omega _p}ωp​ 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 ≈\approx≈ 4 ×\times× 1027 m-3. Taking ε0{{\varepsilon _0}}ε0​= 10- 11 and m ≈\approx≈ 10- 30, where these quantities are in proper SI units.
  1. A
    800 nm
  2. B
    600 nm
  3. C
    300 nm
  4. D
    200 nm
View written solutionFree

Correct answer: B

  1. Plasma frequency formula

For free electrons oscillating against fixed positive ions, the plasma angular frequency is

ωp=Ne2mε0.\omega_p = \sqrt{\frac{N e^2}{m\varepsilon_0}}.ωp​=mε0​Ne2​​.

Plasma reflection occurs when the incident electromagnetic wave has frequency near this value, so

ω≈ωp.\omega \approx \omega_p.ω≈ωp​.

The corresponding wavelength is

λ=2πcωp.\lambda = \frac{2\pi c}{\omega_p}.λ=ωp​2πc​.
  1. Substitute the given approximate values

Given:

N=4×1027 m−3,e=1.6×10−19 C,N = 4\times 10^{27}\,\text{m}^{-3},\qquad e = 1.6\times 10^{-19}\,\text{C},N=4×1027m−3,e=1.6×10−19C, m≈10−30 kg,ε0≈10−11.m \approx 10^{-30}\,\text{kg},\qquad \varepsilon_0 \approx 10^{-11}.m≈10−30kg,ε0​≈10−11.

Now,

e2=(1.6×10−19)2=2.56×10−38.e^2 = (1.6\times 10^{-19})^2 = 2.56\times 10^{-38}.e2=(1.6×10−19)2=2.56×10−38.

So,

Ne2=(4×1027)(2.56×10−38)=10.24×10−11=1.024×10−10.N e^2 = (4\times 10^{27})(2.56\times 10^{-38}) = 10.24\times 10^{-11} = 1.024\times 10^{-10}.Ne2=(4×1027)(2.56×10−38)=10.24×10−11=1.024×10−10.

Also,

mε0=(10−30)(10−11)=10−41.m\varepsilon_0 = (10^{-30})(10^{-11}) = 10^{-41}.mε0​=(10−30)(10−11)=10−41.

Hence,

Ne2mε0=1.024×10−1010−41=1.024×1031.\frac{N e^2}{m\varepsilon_0} = \frac{1.024\times 10^{-10}}{10^{-41}} = 1.024\times 10^{31}.mε0​Ne2​=10−411.024×10−10​=1.024×1031.

Therefore,

ωp=1.024×1031≈3.2×1015 rad/s.\omega_p = \sqrt{1.024\times 10^{31}} \approx 3.2\times 10^{15}\,\text{rad/s}.ωp​=1.024×1031​≈3.2×1015rad/s.
  1. Find the wavelength

Using

λ=2πcωp,\lambda = \frac{2\pi c}{\omega_p},λ=ωp​2πc​,

with c=3×108 m/sc = 3\times 10^8\,\text{m/s}c=3×108m/s,

λ=2π(3×108)3.2×1015.\lambda = \frac{2\pi (3\times 10^8)}{3.2\times 10^{15}}.λ=3.2×10152π(3×108)​. λ≈18.85×1083.2×1015=5.89×10−7 m.\lambda \approx \frac{18.85\times 10^8}{3.2\times 10^{15}} = 5.89\times 10^{-7}\,\text{m}.λ≈3.2×101518.85×108​=5.89×10−7m. λ≈589 nm≈600 nm.\lambda \approx 589\,\text{nm} \approx 600\,\text{nm}.λ≈589nm≈600nm.
  1. Match with the options

The closest option is:

600 nm\boxed{600\,\text{nm}}600nm​

So the correct option is B.

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