JEE Advanced · Physics · 16. Waves & Sound
Paragraph :
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 being 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 \(\omega\), 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 high reflectivity of metals.
Question :
Estimate the wavelength at which plasma reflection will occur for a metal having the density of electrons \(N \approx 4 \times 10^{27} \mathrm{~m}^{-1}\). Take \(\varepsilon_0=10^{-11}\) and \(m=10^{-30}\), where these quantities are in proper SI units.
- A \(800 \mathrm{~nm}\)
- B \(600 \mathrm{~nm}\)
- C \(300 \mathrm{~nm}\)
- D \(200 \mathrm{~nm}\)
Answer & Solution
Correct Answer
(B) \(600 \mathrm{~nm}\)
Step-by-step Solution
Detailed explanation
\(
\begin{gathered}
\omega=2 \pi f=\frac{2 \pi c}{\lambda} \\
\therefore \lambda=\frac{2 \pi c}{\omega}=\frac{2 \pi c}{\sqrt{N e^2 / m \varepsilon_0}}
\end{gathered}
\)
Substituting the values, we get
\(
\lambda=600 \mathrm{~nm}
\)
\(\therefore\) Correct option is (b).
Analysis of Question
(i) Paragraph does not make the solution very clear.
(ii) Solution is simple but based on error and trial method.
\begin{gathered}
\omega=2 \pi f=\frac{2 \pi c}{\lambda} \\
\therefore \lambda=\frac{2 \pi c}{\omega}=\frac{2 \pi c}{\sqrt{N e^2 / m \varepsilon_0}}
\end{gathered}
\)
Substituting the values, we get
\(
\lambda=600 \mathrm{~nm}
\)
\(\therefore\) Correct option is (b).
Analysis of Question
(i) Paragraph does not make the solution very clear.
(ii) Solution is simple but based on error and trial method.
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