Question

Compute lp and N, for the following cases: (a) A glow discharge, with n = 1010 cm-3, KT, = 2 eV. (b) The earth's ionosphere, with n= 106 cm-3, KT, = 0.1 eV. (c) A 6-pinch, with n=1017 cm-3, T. = 800 eV.

Answer 1

(a)
`L_(D) = 1.052* 10^(- 4) m`

N =
`4.87* 10^(4)`

(b)
`L'_(D) = 5.531* 10^(- 6) m`

N' =
`7.087* 10^(- 4)`

(c)
`L''_(D) = 4.43* 10^(- 13) m`

N'' =
`3.63* 10^(- 14)`

Solution:

As per the question, we have to calculate the Debye,
`L_(D)` length and N for the given cases.

Also, we utilize the two relations:

1.
`L_(D) = \sqrt{(KT\epsilon_(o))/(ne^(2))}`

2. N =
`(4)/(3)n\pi(L_(D))^(3)`

Now,

(a) n =
`10^(10) cm^(- 3)* (10^(- 2))^(- 3) = 10^(16) m^(- 3)`

KT = 2 eV

Then

`L_(D) = \sqrt{(2* 1.6* 10^(- 19)* 8.85* 10^(- 12))/(10^(16)(1.6* 10^(- 19))^(2))}`

(Since,

e =
`1.6* 10^(- 19) C`

`\epsilon_(o) = 8.85* 10^(- 12) F/m`)

Thus

`L_(D) = 1.052* 10^(- 4) m`

Now,

N =
`(4)/(3)* 10^(16)\pi(1.052* 10^(- 4))^(3) = 4.87* 10^(4)`

(b) n =
`10^(6) cm^(- 3)* (10^(- 2))^(- 3) = 10^(12) m^(- 3)`

KT = 0.1 eV

Then

`L'_(D) = \sqrt{(0.1* 1.6* 10^(- 19)* 8.85* 10^(- 12))/(10^(12)(1.6* 10^(- 19))^(2))}`

`L'_(D) = 5.531* 10^(- 6) m`

N' =
`(4)/(3)* 10^(12)\pi(5.531* 10^(- 6))^(3) = 7.087* 10^(- 4)`

(c) n =
`10^(17) cm^(- 3)* (10^(- 2))^(- 3) = 10^(23) m^(- 3)`

KT = 800 eV

`L''_(D) = \sqrt{(800* 1.6* 10^(- 19)* 8.85* 10^(- 12))/(10^(23)(1.6* 10^(- 19))^(2))}`

`L''_(D) = 4.43* 10^(- 13) m`

N'' =
`(4)/(3)* 10^(23)\pi(4.43* 10^(- 13))^(3) = 3.63* 10^(- 14)`