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IIT JEE Advanced 2011 Paper 1

Section: Physics

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Question : 13 of 69

Marks: +1, -0

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

\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, where a part of the energy is absorbed and a part of it is reflected. As ω approaches

\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 ×

10^{27} \mathrm{m}^{-3}

\varepsilon_0

10^{-11}

10^{-30}

, where these quantities are in proper SI units.

[JEE Adv 2011 P1]

800 nm
600 nm
300 nm
200 nm

Solution:

We have ω =

\sqrt{\frac{Ne^2}{m\varepsilon_0}}

; N ~ 4 ×

10^{-27} \mathrm{m}^{-3}

;

\varepsilon_0

=

10^{-11} \mathrm{C}^2\mathrm{N}^{-1}\mathrm{m}^{-2}

; m =

10^{-30} \mathrm{kg}

.

ω =

\sqrt{\frac{4 \times 10^{27} \times (1.6 \times 10^{-19})^2}{10^{-30} \times 10^{-11}}}

= 3.2 ×

10^{15} \mathrm{Hz}

Therefore,

ω =

\sqrt{\frac{4 \times 10^{27} \times (1.6 \times 10^{-19})^2}{10^{-30} \times 10^{-11}}}

= 3.2 ×

10^{5} \mathrm{Hz}

Thus, we get the wavelength

λ =

\frac{2\pi c}{\omega}

=

\frac{2 \times 3.14 \times 3 \times 10^8}{3.2 \times 10^{15}}

~ 600 nm

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