Question

Estimate the wavelength at which plasma reflection will occur for a metal having the density of electrons $\text{N}\approx \text{4}\times \text{1}{{\text{0}}^{27}}\ {{\text{m}}^{-3}}$. Take ${{\varepsilon }_{o}}\approx {{10}^{-11}}$ and $m\approx {{10}^{-30}}$, where these quantities are in proper SI units
A. $800\ \text{nm}$
B. $600\ \text{nm}$
C. $300\ \text{nm}$
D. $200\ \text{nm}$

Answer 1

Determine the plasma frequency using its formula. Convert the angular frequency to frequency and determine the frequency. Determine the wavelength using the obtained frequency.
Formula Used:
${{\omega }_{p}}=\sqrt{\dfrac{\text{N}{{\text{e}}^{2}}}{m{{\varepsilon }_{0}}}}$
$\omega =\dfrac{2\text{ }\\!\\!\pi\\!\\!\text{ c}}{\lambda }$

Complete answer:
Neutral plasma consists of a dense collection of equal number of electrons and positive ions. Some solids can also be termed as neutral plasma if they contain fixed positive ions surrounded by fixed electrons. When an external electric field is applied, the electrons are displaced relatively from the positive ions with a natural angular frequency. These electrons start oscillating with a natural angular frequency known as plasma frequency $\left( {{\omega }_{p}} \right)$ when the external electric field becomes zero. To sustain the oscillations, a time varying electric field needs to be applied that has an angular frequency $\omega$, where a part of the energy is absorbed and a part of it is reflected. The resonance condition is reached when $\omega$ becomes equal to ${{\omega }_{p}}$ and all the energy is reflected.
The plasma frequency is calculated as,
${{\omega }_{p}}=\sqrt{\dfrac{\text{N}{{\text{e}}^{2}}}{m{{\varepsilon }_{0}}}}$
Here, $\text{N}$ is the electron number density, $\text{e}$ is electronic charge, m is mass of electron and ${{\varepsilon }_{0}}$ is permittivity of vacuum.
Substitute the given values and calculate the plasma frequency.
$\begin{aligned} & {{\omega }_{p}}=\sqrt{\dfrac{\text{N}{{\text{e}}^{2}}}{m{{\varepsilon }_{0}}}} \\\ & =\sqrt{\dfrac{4\times {{10}^{27}}\times {{\left( 1.6\times {{10}^{-19}} \right)}^{2}}}{{{10}^{-30}}\times {{10}^{-11}}}} \\\ & =1.6\times 2\times {{10}^{15}} \\\ & =3.2\times {{10}^{15}} \end{aligned}$
Now, angular frequency and frequency are calculated as,
$\omega =2\text{ }\\!\\!\pi\\!\\!\text{ }f$ …… (1)
Now, frequency and wavelength are related as,
$c=f\lambda$
Here, $c$ is the speed of light and $\lambda$ is wavelength.
Use the above relation in equation (1) and determine the value of wavelength. The calculations can be seen as,

& \lambda =\dfrac{2\text{ }\\!\\!\pi\\!\\!\text{ c}}{{{\omega }_{p}}} \\\ & =\dfrac{2\times 3.14\times 3\times {{10}^{8}}}{3.2\times {{10}^{15}}} \\\ & \approx 600\ \text{nm} \end{aligned}$$ Thus, the plasma reflection will occur at $600\ \text{nm}$. **Thus, option (B) is correct.** **Note:** The plasma frequency has to be converted in frequency to determine the wavelength at which reflection occurs. Use SI units throughout to obtain the correct answer. Use the mass of the electron as provided specifically in the question.