Monday, April 20, 2015

Astrophysics Solutions (3)


1)Consider the ionization of helium, with three stages of ionization for which the partition functions are:    Z I =    Z 3    =   1,  and Z 2 =    2

With ionization energies:


 c  1 =   24.58 eV    and c  2 =   13.6 eV  

For a temperature of T = 5040 K write out two versions of the Saha equation for:
 

log( P e N 2 /N 1)   and log( P e N 3 /N 2)  
 

And hence, compute:   (P e N 2 /N 1)   and (P e N 3 /N 2)  

 
 Solution: Using the information given and the Saha eqn. we may write:

a)     log( P e N 2 /N 1)   = -0.48  + log (2 Z i + 1 / Z i ) + 2.5 log T – 5040 c i/ T

 

 = -0.48  + log (2 Z 2 / Z 1 ) + 2.5 log T – 5040 c 1 / T

 

=  -0.48  + log (2 (2) / 1 ) + 2.5 log (5040)  – 5040 (24.85  eV ) / 5040

 

b)     log( P e N 3 /N 2)   =  - 0.48  + log (2 Z 3 / Z 2 ) + 2.5 log T – 5040 c 2 / T

 

= - 0.48  + log (2 (1) / 2 ) + 2.5 log (5040) – 5040 (13.6 ) / 5040

 
Now, simplify each equation and solve for the respective quantities:

(P e N 2 /N 1)   and (P e N 3 /N 2)  

 
a) log( P e N 2 /N 1)   = -0.48 +  log(2) + 9.25 -   24.85

 

=   -0.48 + 0.30 + 9.25 – 24.85 =   -15.78

 

(P e N 2 /N 1)   =   Antilog(-15.78)  =   1.6 x 10  -16

 

b) log( P e N 3 /N 2)   =  - 0.48  + log (1) + 9.25  - 13.6


=   -0.48 + 0  + 9.25  - 13.6   =   - 4.83


(P e N 3 /N 2)   =  Antilog(-4.83)  =  1.5 x 10 -5    Ans.

 

2)The diagram below was sketched by a first year astrophysics student. It represents part of the emission spectrum of atomic hydrogen. It contains a series of lines, and the wavelengths of some (in nm) are marked. There are no lines in the series less than 91.2 nm


(a)   In which region of the E-M spectrum would these lines occur?

 
(b) Obtain a relation between E and l   and thence find the photon energies equivalent for all the wavelengths marked.


(c)Use this information to map a partial energy level diagram for hydrogen. Show and label clearly the electronic transitions responsible for the emission lines labeled.

 

Solutions: 
 
a)the lines occur in the UV (ultraviolet) part of the EM spectrum (Lyman series)

 
b) We have:  El  = 1.99 x 10 -16  J nm 

Then the energy for the given wavelength can be obtained from:

 
E  = 1.99 x 10 -16  J nm  / (l  nm)

 
The energies for each wavelength, starting from the shortest, are then:

E1 = 2.18 x 10 -18   J


E2 = 2.09 x 10 -18   J

 
 E3 = 2.04 x 10 -18   J


 E4 = 1.93 x 10 -18   J


E5 = 1.63 x 10 -18   J

For each of the above, an energy difference (i.e. between energy levels) can be confirmed such that the wavelength of the photon emitted can be found. Since E = hf = h (c/ l):

l =   hc/ (E2 – E1) 

Here: E(n=2) =  - 13.6/4  (eV) =  -(3.4) x 1.6 x 10-19 J  =

 -5.4 x  10-19 J 

E(n=1) =  -(13.6)  x 1.6 x 10-19 J  = -21.8 x 10-19 J 

 
Then the energy difference is:


E2 – E1 = [- 5.4 – (-21.8)]  x 10-19 J  = 16.4 x 10 -19 J 

l =   hc/ (E2 – E1) 

l =    (6.626 x 10- 34 J-s)(3 x 10 8 m/s)/ 16.3 x 10-19 J 

 
l =    1.21 x 10 -7 m 

Which is the 121.7 nm value.


In similar fashion all the other wavelengths can be computed using the energy difference, so that:

 l4 =   hc/ E4  =   102.7 nm

l3 =   hc/ E3  =   97.3 nm

l2 =   hc/ E2  =   95.0 nm

 
l1 =   hc/ E1  =   91.2  nm

The sketch of the line transitions is shown below with the shortest wavelength (l1 ) corresponding to the largest energy level difference, i.e. E1 =  (n5 – n1).

 
3)a) A stellar atmosphere is composed of pure hydrogen, at a temperature of 9600 K and with electron pressure, P e    =  200 dyne cm -2 .  Use the Saha equation to compute the fraction of atoms that are ionized in relation to the total, i.e. find:   N II /( N I   + N II )  
 
The partition functions are:  Z II =   1 and  Z I =   2
 
Hint: Check to see whether (E 2  -  E 1 ) >> kT with
 
k =  8.6174 x 10 -5 eV/K
 
So the Boltzman factor  exp (- E2 – E1) / kT<< 1
 

a)Solution:

Note that (E2 – E1) = 10.2 eV and 10.2 eV >> kT

Where kT=  (8.62 x 10 -5 eV/K) (9600K) = 0.82  eV

Then the Boltzmann factor:

exp (- E2 – E1) / kT is much less than  1

Then:

(N i + 1 /N i)  =   (kT)/ P e  [2 π  me kT/ h2 ]3/2  ( e - c i/ kT)

= (  N II )/ N I   

e - c i/ kT   =     exp (-13.6/ 0.82 eV) =  6.2 x 10 -8

Then we find:

(kT)/ 200 dyne cm -2 )[ 2 π  ((9.1 x 10  -28 g) (0.82)/ (6.62 x 10  -27 erg-s  )2 ]3/2  ( e - c i/ kT)


= 
0.933

But:  N II / N total       =

N II /( N I   + N II )   =  { N II / N I    / (1 +  N II / N I )  }

=  [0.933 /  1 + 0.933]  = 0.933/ 1.933 =  0.482

Or nearly half of hydrogen is ionized at 9600K

For (b) 0ne will find by comparing the computed fractions for ionization at the other temperatures, the one for 9600K is at a higher proportion for ionized fraction, thus showing max. intensity is obtained near that temp. (actually at 9520K which can be verified by computation)

(4) Calculate how far you could see through the Earth’s atmosphere if it had the same opacity as the solar atmosphere, where:

k l   =   0.264 cm 2 /g.

Take the density of Earth’s atmosphere as:   

r   =   0.0012 g cm -3

The density of the solar atmosphere is:

r s   =  2.5 x 10  -7 g cm -3

And we know the intensity declines over a characteristic distance (length scale):

  =   1/ k l r

Solution:

We can use the last relation:    =   1/ k l r

But substitute the solar opacity:  k l   =   0.264 cm 2 /g

So: 

  =   1/   (0.264 cm 2 /g) (0.0012 g cm -3 )  =  3.1 x 10  3 cm   =  

3.1 x 10  1 m   = 31m

Earth’s opacity is normally:  k e   =   0.0001 cm 2 /g

So one would theoretically be able to see:

  =   1/ k e  r   =  8.3 x 10  6 cm   =   8.3 x 10  4 m   or 83 km

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