| Mean speed of D2 molecules at 293 º K:
½ . m . v2
=
3/2
.
k . T
(see:
hyperphysics kinetic temperature )
k =
Bolzmann constant
= 1,38 E-23 J/K ( k = R/Na,
R=8,31 Nm/mol, K
Na = 6,0 E23)
m = massa D2 molecule =
2.
2. 1,67E-27 = 6,68E-27 kg
v2
=
3/2
. 1,38 E-23 .293 / (
½ . 6,68E-27 )
v = 1,35 . 103
m/s
-> kinetic energy = ½ . 6,68E-27 . (1,35 . 103
) 2
= 6,09E-21 J => 6,09E-21 /(1,6E-19) = 0,038 eV
|
|
v
= 3. 104
m/s of D+ ions.
m(D+
ion) =
2. 1,67. 10-27 kg = 3,34 . 10-27
kg
½ . m . v2
=
3/2
.
k . T
-> T =
⅓ . m . v2
/ k = ⅓ . 3,34 . 10-27 . (3.104
)2 / (1,38 .
10-23) =
7,26 . 104 K
½ . m . v2
=
ΔV. q (ΔV = voltage
difference needed to accelerate the ions; q = charge of a
ion = 1,6 . 10-19 C )
ΔV = ½ . m . v2
/ q =
½ . 3,34 . 10-27 . ( 3. 104
)2 / (1,6 . 10-19 )
=
9,4 V
|
|
v
= 3. 104
m/s of electrons
m(electron)
=
9,1. 10-31 kg
ΔV = ½ . m . v2
/ q =
½ . 9,1. 10-31 . ( 3. 104
)2 / (1,6 . 10-19 )
= 0,0026 V
|
|
v
= 3. 105
m/s of D+ ions.
-> T =
⅓ . m . v2
/ k = ⅓ . 3,34 . 10-27 . (3.105
)2 / (1,38 .
10-23) =
7,26 . 106 K
½ . m . v2
=
ΔV. q
ΔV = ½ . m . v2
/ q =
½ . 3,34 . 10-27 . ( 3. 105
)2 / (1,6 . 10-19 )
= 900,4 V
> kinetic energy = ½ . 3,34E-27 . (3 . 105
)2
= 1,5E-16 J => 1,5E-16 /(1,6E-19) = 939 eV
|
|
v
= 3. 105
m/s of electrons
ΔV = ½ . m . v2
/ q =
½ . 9,1. 10-31 . ( 3. 105
)2 / (1,6 . 10-19 )
= 0,26 V
|
|
v
= 3. 106
m/s of D+ ions.
-> T =
⅓ . m . v2
/ k = ⅓ . 3,34 . 10-27 . (3.106
)2 / (1,38
.
10-23) =
7,26 . 108 K
½ . m . v2
=
ΔV. q
ΔV = ½ . m . v2
/ q =
½ . 3,34 . 10-27 . ( 3. 106
)2 / (1,6 . 10-19 )
= 90kV
> kinetic energy = ½ . 3,34E-27 . (3 . 106
)2
= 1,5E-14 J => 1,5E-14 /(1,6E-19) = 93,9 keV |
|
Some
speed - temperature - voltage calculations
20 February
2014 ..
by
Rinze
Joustra www.valgetal.com