1)

A dense collection of an 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. lf the electric field becomes zero, the electrons being to oscillate about the positive ions with a natural! angular frequency 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, where a part of the energy is absorbed and a part of it is reflected. As $\omega$  approaches $\omega_{p}$,  all the free electrons are set to resonance together and all the energy is reflected. This is the explanation of the high reflectivity of metals

Estimate the wavelength at which plasma reflection will occur for a metal having the density of electrons $N= 4 \times 10^{27}m^{-1}$. Take  $\epsilon_{0}=10^{-11}$ and $m=10^{-30}$ where these quantities are in proper SI unit


A) 800 nm

B) 600 nm

C) 300 nm

D) 200 nm

Answer:

Option B

Explanation:

 $\omega =2 \pi f =\frac{2 \pi c}{\lambda}$

 $\therefore$  $\lambda = \frac{2 \pi c}{ \omega}= \frac{2 \pi c }{\sqrt{Ne^{2}/m\epsilon_{0}}}$

 Substituting  the values , we get

 $\lambda= 600nm$

 $\therefore$ correct option is (b)