Application Center - Maplesoft

App Preview:

Calculating the Amount of Hydrogen on the Sun's surface

You can switch back to the summary page by clicking here.

Learn about Maple
Download Application




``Calculating the Amount of Hydrogen on the Sun's surface ***

 

restart

``

Problem: (A) Use two lines of the Paschen series of hydrogen, along with the general curve of growth of the Sun, to calculate the number of hydrogen atoms with electrons in the n = 3 orbital above each square metre of the Sun's surface. The Paschen series results from an electron making an upward transition from the n = 3 orbital of the hydrogen atom. (B) Then use the Bolztmann and Saha equations to calculate the total number of hydrogen atoms above each square metre of the Sun's surface.

 

Hints:

 

Use the two Paschen lines given in the table and calulate log10(W/λ).

Find the value of log10(f (λ/500 nm) for each of the two lines.

Find log10 N[a] (where N[a] = the number of atoms) for each of the two lines.

Find the average log10 N[a] and solve for N[a].

Use the Boltzmann equation to find the ratios of the numbers of atoms in the three levels.

From this result, calculate the total number of hydrogen atoms per square metre at the solar surface.

Use the Saha equation to check the ionization state of the atoms to determine whether or not the Boltzmann result can be accepted.    

 

Data:

 

The following table, from Aller (1971), gives the wavelength (λ), line width (W), and oscillator strength (f) of the Paschen gamma and Paschen delta lines of hydrogen in the Sun.

 

lambda*nm

W*nm

f

1093.8*[`Paγ`]

.22

0.554e-1

1004.9*[`Paδ`]

.16

0.269e-1

 

The following is the curve of growth of the Sun, from Aller (1971).

 

knvrt := convert(1, 'units', 'eV', 'J')

0.1602176462e-18

(1)

``

Energies for the n = 3, n = 2, and n = 1 states, respectively:

NULL

E[3] := -1.51

-1.51

(2)

E[2] := -3.40

-3.40

(3)

E[1] := -13.6

-13.6

(4)

T := 5800

5800

(5)

P[e] := 1

1

(6)

Partition numbers for levels I and II:

 

Z[I] := 2

2

(7)

Z[II] := 1

1

(8)

````

chi := 13.6*knvrt

0.2178959988e-17

(9)

NULLNULLNULL

with(ScientificConstants) 

 

k := evalf(Constant(k))

0.1380650277e-22

(10)

m[e] := evalf(Constant(m[e]))

0.9109381882e-30

(11)

h := evalf(Constant(h))

0.662606876e-33

(12)

 

 

Useful Equations:

 

log10(f (λ/500 nm) = log10(f  * λ/500)

 

log10 Na =  log10[f Na(λ/500 nm)] -  log10(f (λ/500 nm)  

 

N[b]/N[a] = g[b]*exp((-E[b]-E[a])/(k*T))/g[a]

``NULL

where N refers to the number of atoms in a given state, g = the number of degenerate states (g=2n2 for hydrogen), E is the energy, k = the Bolzmann constant, and T is the temperature in kelvins.

 

N[i+1]/N[i] = 2*k*T*Z[i+1]*(2*Pi*m[e]*k*T/h^2)^(3/2)*exp(-chi[i]/(k*T))/(P[e]*Z[i])

 

where the pressure (Pe) = 1 N m-2, ZI = 2, ZII = 1, and χi = 13.6 eV.

 

 

 

Solution (A):

 

For both lines, calculate log10(W/λ).

 

log[10](.22/(1093.8))

-3.696515238

(13)

for Paγ, and

NULL

log[10](.16/(1004.9))

-3.798002863

(14)

for Paδ.

 

Trace these figures on the solar curve of growth to find corresponding values for log10[f Na(λ/500 nm)]. The results are 19.21 for the 1093.8 nm line and 18.96 for the 1004.9 nm line. Next, the value of log10(f (λ/500 nm) is calculated for each line.

 

log[10]((1/500)*(0.554e-1*1093.8))

-.9165223206

(15)

for Paγ, and

 

log[10]((1/500)*(0.269e-1*1004.9))

-1.267094878

(16)

for Paδ.

 

 

Find log10 na for both lines:

 

19.21+.9165223206

20.12652232

(17)

NULL

for Paγ, and

 

18.96+1.267094878

20.22709488

(18)

for Paδ.

 

 

The average of these values is

 

(1/2)*(20.12652232+20.22709488)

20.17680860

(19)

Therefore,

 

log[10](N[a]) = 20.18

ln(N[a])/ln(10) = 20.18

(20)

N[a] := solve(log[10](N[a]) = 20.18, N[a])

0.1513561248e21

(21)

NULL

According to the solar curve of growth, there are approximately 1.51 * 1020 atoms of hydrogen with electrons in the n = 3 orbital above every square metre of the Sun's surface.

 

 

Solution (B) To find the total number of hydrogen atoms above every square metre of the Sun's surface, the Boltzmann and Saha equations can be used. Using Boltzmann's equation, the ratio of the number of hydrogen atoms in the third state to that in the second state is

NULL

(18*(1/8))*exp(-(E[3]-E[2])*knvrt/(k*T))

0.5127570812e-1

(22)

NULL

and the ratio of the number of hydrogen atoms in the third state to that in the first state is

``

 

(18*(1/2))*exp(-(E[3]-E[1])*knvrt/(k*T))

0.2811399852e-9

(23)

``

Therefore, for every neutral hydrogen atom in n = 3, there are

NULL

1/0.5127570812e-1

19.50241229

(24)

NULL

in the n = 2 state and

 

1/(2.811399852*10^(-10))

3556946904.

(25)

NULL

in the n = 1 state. Using the solar curve of growth, this would give a total of

NULL

N[a]*(1+19.50241229+3.556946904*10^9)

0.5383657027e30

(26)

hydrogen atoms per square metre of the Sun's surface.

 

 

The ratio of ionized to neutral atoms according to the Saha equation is

 

 

2*k*T*Z[II]*(2*Pi*m[e]*k*T/h^2)^(3/2)*exp(-chi/(k*T))/(P[e]*Z[I])

0.1651490286e-4*2^(1/2)*Pi^(3/2)

(27)

"(->)"

0.13006e-3

(28)

``

``

This shows that almost all of the hydrogen atoms are neutral. Therefore, the result of the Boltzmann equation can be accepted.

 

N[total] = 5.383657027*10^29/Unit('m')^2

N[total] = 0.5383657027e30/Units:-Unit('m')^2

(29)

``

-----------------------------------------------------------------------------------

Reference

 

Aller, L. (1971). Atoms, Stars, and Nebulae. (Rev. Ed.). Cambridge, MA: Harvard University Press.

 

 

 

``