Ge/Ch 128 -- Cosmochemistry
Solutions to problem set #1
1. Can the existence of diatomic molecules in the interstellar medium be explained by the
disruption of larger grains alone (from stellar mass loss)?
We want to compare the observed mass input (from stellar mass loss) with the observed
diatomic molecule abundance (predominantly H2):
INPUT = 4x10-10 [Msun yr-1 pc-2] * 1.98x1033 [g/Msun] * 1/30 [moles/g]
* 6.022x1023 [molec/mol] *  Rgalaxy2 [pc2]
and with Rgalaxy = 2x104pc ==> INPUT = 2.01x1055 [molec/yr]
Now we calculate the loss rate of for the observed diatomic abundances:
Observed # density of H atoms: 1cm-3
Observed # density of diatomics: 10-7 * 1 cm-3 = 10-7 cm-3
So the total number of diatomics observed in diffuse gas in the galaxy is:
Nobserved = 10-7 [cm-3] *  Rgalaxy2 [pc2] * hgalaxy [pc] * 300 pc
*3.09x1018 [cm/pc]
= 1.11x1060 molecules in galaxy (observed)
For a dissociation rate of 1.11x10-10 sec-1 =  ==>  =1/ ~ 300 yrs
so the average lifetime of a diatomic is ~300yrs
==> equilibrium destruction rate = OBSERVED LOSS = Nobserved /destruction
= 3.69x1057 molec/yr
To balance this observed loss, we would have to input molecules at the same rate. But
the observed input rate from stellar mass loss is only:
OBSERVED INPUT
REQUIRED INPUT
=
2.01x1055 [molec/yr] = 0.005 (0.5% of the
3.69x1057 [molec/yr]
required amt.)
So other sources are necessary to produce molecules in the interstellar medium, i.e. we
must make bonds.
2. We are given:
d = nHI e2/me c * (v) * gnf * dL
(Note that this expression does not include the oscillator strength. This parameter
depends on the particular atom or molecule involved, and for HI it is ~0.41. The above
equation will give numbers ~2 times reality because of this omission.)
The total optical depth is = L d
-if we assume all factors are constant over the pathlength:
 = 1 = nHI e2/me c * (v) * gnf * L ==> L= me c / (nHI e2 (v) * gnf )
Tau
freq(L)
Freq---->
This curve drops mostly due to (v) (a little due to gnf ).
What is the path length at 912 Å and 180 Å?
912 Å
 = 3.29x1015 Hz
gnf = 0.797 (Spitzer, p.106)
==> L (912 Å) = 9.1x10-28 * 3x1010 * 3.29x1015 = 7.778 x 1016 cm
1*2 (4.8x10-10)2 *0.797
using 1 pc = 3.08x1018 cm
==> L(912 Å, =1) = 0.025 pc
150 Å
 = 1.67x1016 Hz
gnf = 0.994 (Spitzer, p.106)
==> L (150 Å) = 9.1x10-28 * 3x1010 * 1.67x1016 = 8.12 x 1018 cm
1*2(4.8x10-10)2 *0.994
==> L(150 Å, =1) = 2.63 pc
Dense clouds are on the order of several parsecs in diameter. The typical pathlength for a
912 Å photon is << cloud size so this transition is very sharp. Viewing a dense cloud at
912Å, we would only see the outer surface. At 180Å, L is on the order of the cloud size
so, viewing a cloud at 180Å, we could see deeper into the cloud.
3.The rate at which atoms stick to a grain surface is just the product of the collision rate
and the sticking probability f(x).
The collision rate is given by:
Rate = (ngrain  a2) * v* natom
where v= thermal velocity
 = ngrain  a2 = total grain cross sectional area
-so the sticking rate is:
Rate = dnatom /dt = -  v natom fatom
==> dnatom /natom = -  v natom fatom dt
==> natom/no (atom) = e- v f t
So we have an expression for the depletion as a function of time:
Depletion is defined as:
D = log10 n/ncosmic
(Class notes, p. 2 caption in table)
= log [ (n/no) * (no/ncosmic) ]
where no/ncosmic = 0.1 (given)
for Ca ++ :
n(Ca++)/no(Ca++) =
where = 1.8x10-21 * 90
v = thermal velocity = 1.29x104 (T/M) ½
(with M in amu)
= 1.29x104 (95/40)½ = 1.99x104 cm/sec
f = 1.0
t = 1015 sec
==> D(Ca++) = log [3.98 x 10-2 (0.1)]
= -2.40
for Si + :
n(Si+)/no(Si+) =
where = 1.8x10-21 * 90
v = 1.99x104 cm/sec
f = 0.25
t = 1015 sec
==> D(Ca++) = log [4.47 x 10-1 (0.1)]
= -1.35
This helps illustrate low how important the sticking probability is.
4.
a) Cross-sectional area of the grain:
r = 1000 Å = 1 x 10-5 cm = grain radius
 r2 = 3.14 x 10-10 cm2
total area that 1037 ergs/sec passes through at 400AU:
1 AU = 1.5 x 1013 cm
4  R2 = 4pi (400*1.5 x 1013)2
R=distance from star to grain
33
2
= 4.5 x 10 cm
 the grain receives : 3.14 x 10-10 (1037) = 6.98 x 10-6 ergs/sec
4.5 x 1032
radiates as a blackbody:
Fout = ( Te4) * surface area
= Te4 (5.67 x 10-5) 4  (1 x 10-5)
= Te4 (7.13 x 10-14)
At equilibrium, Power in = Power out
==> 6.98x10-6 ergs/sec = Te4 (7.13 x 10-14)
==> Te = 99.5 K
(H2O begins subliming in ultra high vacuum at ~130K)
More generally: Power in = Power out
Pin = Lsun (r2/ 4R2) , where r = grain radius, R = distance from star to grain
Pout =  Te4 4r2




Lsun (r2/ 4R2) =  Te4 4r2
T = (Lsun / 16 R2) (¼
)
or
T  (1/R) (½
also note that the temperature does not depend on grain size!
)
b) If the grain is silicate in composition, c ~ 3 cal/gmK and  ~ 3 gm/cm3
4.
The temperature rise is just: T = E / cM
where E = energy input
c = heat capacity
m = mass


 T = E / (c  4/3 pi r3)
for a 2000 Å grain:



T =
10 eV * 3.83 x 10-20 cal/eV
3 cal/gmK *3 gm/cm3 * 4/3 pi (1000x10-8cm)3
=
1.02 x 10-20
(1000 x 10-8cm)3
=
1.02 x 10-5 K
for a 200 Å grain:
T =
1.02 x 10-20
(100 x 10-8cm)3
=
1.02 x 10-2 K
for a 20Å grain:
T =
1.02 x 10-20
(10 x 10-8cm)3
=
10.2 K
So you can see the heating is not significant for "typical" photons (< 13.6 eV) absorbing,
even for the smallest grains. Cosmic rays (with energies often 103 - 106 eV) can do a lot
more damage.
5.From the Lecture notes, we know that:
bc = (4q2 / vo2) ¼
so:
r = q/L (2 )½
bc 4 = 4q2 / vo2
and rc2 = 2q2
vo 2bc2




L= vo bc
rc2 = (q/L)2 (2 )
=
2q2 = 4q2
vo 2bc2
vo 2
 bc = 2½ rc
x
1
bc2
=
bc2
2
6.The Langevin rate constant for ion-molecule reactions goes as:
k = 2 q (/ )½ cm3 / sec
where q=4.803 x 10-10 csu, alpha = 1.76 x 10-24 cm3
 = reduced mass = m1m2 / ( m1 + m2 ) ~ m1 2/ 2m1 = m1 / 2 = 14.003 amu
 = 1.66 x 10-24 g / amu * 14.003 amu = 2.33 x 10-23 g
so: k = (4.803x10-10 ) * (1.76x10-24 / 2.33x10-23 )½ = 8.29x10-10 [esu/cm-(3/2) g-1]
units?
1esu-cm = (erg - cm3)½ ==> esu = (erg-cm)½
1erg = 1 g cm2 s-2 ==> 1 esu = (g cm3 / s2)½


esu
=
-3/2 ½
cm g


Barrier height?
From notes,
g½ cm3/2 cm3/2
s g½
= cm3
sec
 k = 8.29x10-10 cm3/sec
Ec = L4 / (8  q2 )
and L4 = mu4 vo4 bc4
also,
bc4 = 4 q2 / ( vo2)
so,
Ec =  q2 vo4 / ( vo2 8  q2 ) = vo2 / 2
(Note that the ½vo2 is just the initial K.E., or the total energy of the system.)
vo = ( 8kT/m) ==> = 8kT/m = 8 * 1.38 x 10-16 * 50 = 3.77x108 cm2 / sec2
* 4.66 x 10-23

Ec = ½ (2.33x10-23 ) *(3.77x108 ) = 4.39 x 10-15 ergs * 7.24 x 1015K/erg = 31.8 K
Now, what is the change in K.E. when we bring N2+ and N2 from an infinite separation to
1.5 Å apart? Note that the collision is head on. --> L=0
K.E.initial = Etotal = K.E.final + P.E.final
 K.E. = K.E.final - K.E.initial = - (P.E.final)
and V(r) = - q2/ 2r4 , so -V(1.5Å) = q2/ 2(1.5Å)4


 K.E. = (1.76 x 10-24)*(4.803 x 10-10)2 = 4.01 x 10-12 ergs
2*(1.5 x 10-8)4
4.01 x 10-12 ergs * 7.24 x 1015 K/erg = 29,000K = 2.5 eV at 1.5 Å
7.
OH+ + CO --> HCO+ + O
Table 3.1 of Duley & Williams : (CO) = 1.95 x 10-24 cm3
MOH = 17/6.02 x 10-23 = 2.82 x 10-27 g/molecule
MCO = 28/6.02 x 10-23 = 4.52 x 10-23 g/molecule
==>  = MOH MCO / (MOH + MCO ) = 1.75 x 10-23 g
so: k = 2 q (/ )½ = 2 (4.8 x 10-10) (1.95x10-24/1.75x10-23)½ = 1.01 x 10-9 cm3/sec
Eqn 3.10 of Duley & Williams gives : kADO = kL [ 1 + (cD/½kT)½ ]
Given that D = 0.1 Debye, D / ½ = 0.1 Debye /(1.95 Å3)½= 0.7 [Debye/Å3/2 ]
-consulting figure 3.2, at reasonable temperatures (150 < T < 650 K), c< 0.01.
The ADO correction term (2nd term in kADO expression) is thus:
using 1 Debye = 10-18 (erg-cm3)½
correction term = (cD/½kT)½ ~ c*0.1 x 10-18/(10-24) (2/10-16 *102)½
~ c*0.1 x 10-18/ (10-12 * 10-7 ) ~ c
and as c < 0.01 << 1, the correction factor to the Langevin rate is very small and can be
neglected.



kADO ~ kL
H3+ + HCN --> HCNH+ + H2
Table 3.1 of Duley & Williams : (HCN) = 2.59x10-24 cm3
MH3 = 3.02/6.02 x 10-23 = 5.02 x 10-24 g/molecule
MHCN = (1+12+14) / 6.02x10-23 = 4.49 x 10-23 g/molecule
==>  = MOH MCO / (MOH + MCO ) = 4.52 x 10-24 g
so: k = 2 q (/ )½ = 2 (4.8 x 10-10) (2.59x10-24/4.52x10-24 )½= 2.28 x 10-9 cm3/sec
Eqn 3.10 of Duley & Williams gives : kADO = kL [ 1 + (cD/½kT)½ ]
Given that D = 3 Debye, D / ½ = 3 Debye / (2.59 Å3)½= 1.86 [Debye/Å3/2 ]
figure 3.2 indicates
c~0.26 at 150K
c~0.23 at 650K
The ADO correction term (2nd term in kADO expression) is thus:
using 1Debye = 10-18 (erg-cm3)½
correction term = (cD/½kT)½ ~ c*3x10-18/(2.59x10-24)½(2/10-16 *102)½
so:
at 150 K, kADO = 3.7 kL
at 650 K, kADO = 2.1 kL
So the correction factor cannot be neglected - i.e., the Langevin rate is not adequate to
describe such a reaction.
8. Using a hard sphere approximation, we can roughly determine a collisional
interaction timescale. We will approximate by saying that one particle is moving at a
thermal velocity and the other is stationary. v ~ 3 x 104 cm/sec
r ~1 Å ~ 1 x 10-8 cm
-so spheres "overlap" over ~3-4 Å or 3-4 x10-8 cm
This "overlap" occurs over a time interval of roughly velocity/distance as:
3x104 cm/sec
3x10-8 cm
~ 10-12 sec is a typical neutral-neutral collision timescale.
What are timescales for molecular vibrations?
vib = (4000 - 2000 cm-1) * (3x10-10 cm/sec) = 1.2x1014 - 6x1013 Hz
==>
vib = 2x10-14 - 8x10-15 sec/vibration
So many vibrations can occur in the span of time it takes for a collision. Collisional
energy can transfer to excited vibrational states because of this.
What about timescales for molecular rotations?

rot= (300 - 3 cm-1) * (3x10-10 cm/sec) = 9x1012 - 6x1010 Hz
==>
rot = 1x10-13 - 1x10-11 sec/rotation
These timescales are about the same as those for a collision. Collisional energy not
always likely to transfer to excited rotational states. This depends on specific reaction.
coll ~ rot > vib