JEE Challenger
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PhyMedAdv 2011 • Paper 1

Plasma Frequency Dimensional Analysis for Electron Density and Permittivity

Comprehension Passage

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

Taking the electronic charge as 'ee' and the permittivity as 'ε0\varepsilon_0', use dimensional analysis to determine the correct expression for ωp\omega_p.

Options

A

Nemε0\sqrt{\frac{Ne}{m\varepsilon_0}}

B

mε0Ne\sqrt{\frac{m\varepsilon_0}{Ne}}

C

Ne2mε0\sqrt{\frac{Ne^2}{m\varepsilon_0}}

Correct
D

mε0Ne2\sqrt{\frac{m\varepsilon_0}{Ne^2}}

Topics & Concepts

Units and MeasurementsSHMElectric Field

Step-by-Step Solution

To find the correct dimensional expression for the plasma frequency ωp\omega_p, we first determine the fundamental dimensions of all given physical quantities in terms of Mass (M\text{M}), Length (L\text{L}), Time (T\text{T}), and Electric Current (I\text{I}):

  1. Angular frequency (ωp\omega_p):
    [ωp]=T−1[\omega_p] = \text{T}^{-1}

  2. Number density of electrons (NN): Number of electrons per unit volume
    [N]=1Volume=L−3[N] = \frac{1}{\text{Volume}} = \text{L}^{-3}

  3. Mass of electron (mm):
    [m]=M[m] = \text{M}

  4. Electronic charge (ee):
    [e]=IT[e] = \text{I}\text{T}

  5. Permittivity of free space (ε0\varepsilon_0):
    From Coulomb's Law, F=14πε0q2r2F = \frac{1}{4\pi\varepsilon_0} \frac{q^2}{r^2}:
    [ε0]=[q]2[F][r]2=(IT)2(MLT−2)(L2)=M−1L−3T4I2[\varepsilon_0] = \frac{[q]^2}{[F][r]^2} = \frac{(\text{I}\text{T})^2}{(\text{M}\text{L}\text{T}^{-2})(\text{L}^2)} = \text{M}^{-1}\text{L}^{-3}\text{T}^4\text{I}^2

Assume the relation between ωp\omega_p and the physical parameters is of the form:
ωp∝Naebmcε0d\omega_p \propto N^a e^b m^c \varepsilon_0^d

Substitute the dimensions into the equation:
T−1=(L−3)a(IT)b(M)c(M−1L−3T4I2)d\text{T}^{-1} = (\text{L}^{-3})^a (\text{I}\text{T})^b (\text{M})^c (\text{M}^{-1}\text{L}^{-3}\text{T}^4\text{I}^2)^d

Group the exponents corresponding to each fundamental dimension:
M0L0T−1I0=Mc−d⋅L−3a−3d⋅Tb+4d⋅Ib+2d\text{M}^0 \text{L}^0 \text{T}^{-1} \text{I}^0 = \text{M}^{c - d} \cdot \text{L}^{-3a - 3d} \cdot \text{T}^{b + 4d} \cdot \text{I}^{b + 2d}

Equating exponents on both sides:

  • For Mass (M\text{M}):
    c−d=0  ⟹  c=dc - d = 0 \implies c = d

  • For Electric Current (I\text{I}):
    b+2d=0  ⟹  b=−2db + 2d = 0 \implies b = -2d

  • For Length (L\text{L}):
    −3a−3d=0  ⟹  a=−d-3a - 3d = 0 \implies a = -d

  • For Time (T\text{T}):
    b+4d=−1b + 4d = -1

Substitute b=−2db = -2d into the time equation:
−2d+4d=−1  ⟹  2d=−1  ⟹  d=−12-2d + 4d = -1 \implies 2d = -1 \implies d = -\frac{1}{2}

Now solve for the remaining exponents:
c=−12c = -\frac{1}{2}
b=−2(−12)=1b = -2\left(-\frac{1}{2}\right) = 1
a=−(−12)=12a = -\left(-\frac{1}{2}\right) = \frac{1}{2}

Substituting these exponents back into the expression for ωp\omega_p:
ωp∝N1/2e1m−1/2ε0−1/2=Ne2mε0\omega_p \propto N^{1/2} e^1 m^{-1/2} \varepsilon_0^{-1/2} = \sqrt{\frac{N e^2}{m \varepsilon_0}}

Thus, the correct expression for ωp\omega_p is Ne2mε0\sqrt{\frac{N e^2}{m \varepsilon_0}}.

Correct Answer: C

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