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san4es73 [151]
3 years ago
6

7.22 Ignoring reflection at the air–water boundary, if the amplitude of a 1 GHz incident wave in air is 20 V/m at the water surf

ace, at what depth will it be down to 1 μV/m? Water has μr = 1, and at 1 GHz, r = 80 and σ = 1 S/m.
Physics
1 answer:
Serga [27]3 years ago
5 0

Answer:

z = 0.8 (approx)

Explanation:

given,

Amplitude of 1 GHz incident wave in air = 20 V/m

Water has,

μr = 1

at 1 GHz, r = 80 and σ = 1 S/m.

depth of water when amplitude is down to  1 μV/m

Intrinsic impedance of air = 120 π  Ω

Intrinsic impedance of  water = \dfrac{120\pi}{\epsilon_r}

Using equation to solve the problem

  E(z) = E_0 e^{-\alpha\ z}

E(z) is the amplitude under water at z depth

E_o is the amplitude of wave on the surface of water

z is the depth under water

\alpha = \dfrac{\sigma}{2}\sqrt{\dfrac{(120\pi)^2}{\Epsilon_r}}

\alpha = \dfrac{1}{2}\sqrt{\dfrac{(120\pi)^2}{80}}

\alpha =21.07\ Np/m

now ,

  1 \times 10^{-6} = 20 e^{-21.07\times z}

  e^{21.07\times z}= 20\times 10^{6}

taking ln both side

21.07 x z = 16.81

z = 0.797

z = 0.8 (approx)

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Now, in order to find the wavelength of the electron \lambda_{e}   with this given kinetic energy (hence momentum), we will use the De Broglie wavelength equation:

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