13. 4 √TT (4) [21 (200²) w²amond/thit do 1 v²c² -mu²/2hat 4 2 (2) ` f(u) = Use fos · x² e [ v² (2017) the This 30 => dx _mu ² / 2ha² du when Lo fend der √π/49" 4 ("f(u) du = 3/1 (2016) VTT when v²e with Umr _mu²/2kgT 2 T ✓ √ VIT (21265)3 4 312 √TT (6) Spav df/d₁² = 0 ✓ ; d = 2/ ( ) { 20² - ² (-20) e @_my//PRET} mu² 4/2₁ ( 10² { 2 - 12/2²2) 0² 21207) 12 ✓ muu Kat de ✓ 17 ( 217 ) ✓ m/2k8+ a₂ mprind e-mat" /2k₂7 (c) (d) [ofluido 4 $ (27) ² (²0 = JTC 3.10 = 3/24 (21²5) " Iu² Use [² x ²e" -4x² L 29² doc ✓ mu ² / 2 bat du u² fluido ✓ (2)² [*e-m/ken dur with a = m 31eam Since Uring = √
13. 4 √TT (4) [21 (200²) w²amond/thit do 1 v²c² -mu²/2hat 4 2 (2) ` f(u) = Use fos · x² e [ v² (2017) the This 30 => dx _mu ² / 2ha² du when Lo fend der √π/49" 4 ("f(u) du = 3/1 (2016) VTT when v²e with Umr _mu²/2kgT 2 T ✓ √ VIT (21265)3 4 312 √TT (6) Spav df/d₁² = 0 ✓ ; d = 2/ ( ) { 20² - ² (-20) e @_my//PRET} mu² 4/2₁ ( 10² { 2 - 12/2²2) 0² 21207) 12 ✓ muu Kat de ✓ 17 ( 217 ) ✓ m/2k8+ a₂ mprind e-mat" /2k₂7 (c) (d) [ofluido 4 $ (27) ² (²0 = JTC 3.10 = 3/24 (21²5) " Iu² Use [² x ²e" -4x² L 29² doc ✓ mu ² / 2 bat du u² fluido ✓ (2)² [*e-m/ken dur with a = m 31eam Since Uring = √
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Could you explain this answer please. I have attatched the question in one picture and the answer in another
![13.
4
f(u)
m
312
-mo
(10) = 2 (2) ²e- / MOT
ve
to
(4) [ #1 (51EST) (²² e
U₂ √ x² e ax dx
So
312
2
✓
4
= // (50) ² [²e-mu/that Juo
2
I've ²0 d
du-
=>
This =0
for fler) dur
_mor² / 21₁3²
312
√π/49"
when
when
with
m
312
斤
(fa) du = = ( ²7 )^. / ( 26 )
m
VTT
4
Umi
-
do
2
=
a =
✓
Spak
df/dv
✓
; df
e
dth 2 () { 20² - ² (-2)}
dv
35
= √ (216T | ³3
2kB
णी
-m
2R₁3
m
mu²
e
3/24 ( 1022 - 10/²3 - 1²/24/1
س لے
m/2KBT
as
mu?
KBT
++2
respond
=0
(c)
<vZ
4
UTT
Find
Use Sox
[de
3
926²
е
=
=
∞
m
3½
(2 KT) ¹² 6
<V>
2. <u>
uf(u) du
4
JT
=> <√²7
dx =
✓
=
Since Urims
4
FIT
Use
745
Wie foarte med de
xe
O
3
2
<V²> = √²°~² fluido ✓
za
3 kam
✓
11₂
3/11 (2125) ✓
m
س )
mu²/212₂T du
<V²7
312
Z
m
T
(2) ³ = ( 2 krije
2 .
with
m
312
(2ľ Love mest de
du
2KBT
3√77
89512 ✓
a
=
/
3KBT
m
a
4 (m ³/2 3√πt (21²1) Sh
VIT
8
9
m
2KBT
C m
212B
T
(e)
ordeving of
general
Flu) 1
e
(f) mu² <<h₂>
_mu ²/27537
Correction
Fraction
is
flu) ~
to
F(~)
prot
3
form of
≈
u's important VV
✓
sketch
already
FF
-MA
ZVT
e Vins
Here
2nd
21,3
up
Sprob / 10
Uprav/10
F(x) = (2) ²² d
m
11
63
du
D
4x10³
3571
3/2
2
mu
ZWBT / but
order
✓
20
3/2
3
4
3/1 (2017) ™H [U²³
~Tiol
O
S
pol= ZRBT
m
10 (2hat)
11₂
37
2
S/M (21ST) ³ ( 100 ) ( 21287) de
✓
[20]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3ef9e66d-4b8f-49ef-955b-9683128fda1f%2Fb802435b-656f-4b1d-a929-ce5955dc1688%2F8e7fkcn_processed.png&w=3840&q=75)
Transcribed Image Text:13.
4
f(u)
m
312
-mo
(10) = 2 (2) ²e- / MOT
ve
to
(4) [ #1 (51EST) (²² e
U₂ √ x² e ax dx
So
312
2
✓
4
= // (50) ² [²e-mu/that Juo
2
I've ²0 d
du-
=>
This =0
for fler) dur
_mor² / 21₁3²
312
√π/49"
when
when
with
m
312
斤
(fa) du = = ( ²7 )^. / ( 26 )
m
VTT
4
Umi
-
do
2
=
a =
✓
Spak
df/dv
✓
; df
e
dth 2 () { 20² - ² (-2)}
dv
35
= √ (216T | ³3
2kB
णी
-m
2R₁3
m
mu²
e
3/24 ( 1022 - 10/²3 - 1²/24/1
س لے
m/2KBT
as
mu?
KBT
++2
respond
=0
(c)
<vZ
4
UTT
Find
Use Sox
[de
3
926²
е
=
=
∞
m
3½
(2 KT) ¹² 6
<V>
2. <u>
uf(u) du
4
JT
=> <√²7
dx =
✓
=
Since Urims
4
FIT
Use
745
Wie foarte med de
xe
O
3
2
<V²> = √²°~² fluido ✓
za
3 kam
✓
11₂
3/11 (2125) ✓
m
س )
mu²/212₂T du
<V²7
312
Z
m
T
(2) ³ = ( 2 krije
2 .
with
m
312
(2ľ Love mest de
du
2KBT
3√77
89512 ✓
a
=
/
3KBT
m
a
4 (m ³/2 3√πt (21²1) Sh
VIT
8
9
m
2KBT
C m
212B
T
(e)
ordeving of
general
Flu) 1
e
(f) mu² <<h₂>
_mu ²/27537
Correction
Fraction
is
flu) ~
to
F(~)
prot
3
form of
≈
u's important VV
✓
sketch
already
FF
-MA
ZVT
e Vins
Here
2nd
21,3
up
Sprob / 10
Uprav/10
F(x) = (2) ²² d
m
11
63
du
D
4x10³
3571
3/2
2
mu
ZWBT / but
order
✓
20
3/2
3
4
3/1 (2017) ™H [U²³
~Tiol
O
S
pol= ZRBT
m
10 (2hat)
11₂
37
2
S/M (21ST) ³ ( 100 ) ( 21287) de
✓
[20]
Transcribed Image Text:13. The Maxwell-Boltzmann distribution for the speeds, v, of gas molecules of mass, m, in
a sample with temperature, T, is given by
f(v):
=
4
√
m
2kBT
(b) Find the most probable speed, Uprob.
(c) Find the average speed, (v).
where kB is Boltzmann's constant.
(a) Check that this distribution is appropriately normalised; namely that it satisfies
[ f(v)dv = 1.
=
3/2
(d) Find the r.m.s. speed, Urms
√(v²).
(e) Draw a sketch of f(v) showing Uprob, (v) and Urms as appropriate.
(f) In the low energy limit of f(v) (i.e., when mv²/2 << kBT), estimate the fraction
of the particles with speeds less than Uprob/10.
In this question you may find the following integrals useful:
1
2²e²dr = 2²¹ ²2²e
5.
x³e
3√
8a5/2¹
v²e-mv²/2kBT
-ar² dx =
x¹e-ar²
[ 2² e ²³ da
=
√√T
4a³/2
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