Before going on to solve a set of additional basic problems in astrophysics, let's recap one based on the image shown in the last instalment, of the globular cluster:

http://brane-space.blogspot.com/2010/11/basic-problems-in-astrophysics-2.html

The caption notes that "

Then solving the appropriate equation for the distance (in parsecs) yields:

d = 10^{(18 + 5)/ 5} pc = 10 ^4.6 pc = 39, 800 pc or about 40 kpc (kilo-parsecs) which is a very reasonable distance for a globular cluster.

Now, some more basic problems:

1) The lines of the Balmer series crowd close together as higher series members are considered. If each line were exactly one A (1 Angstrom) wide, how many Balmer lines would be individually visible without overlapping other lines?

Solution:

The Balmer lines are defined according to:

1/ L = R (1/ 4 – 1/n^2)

where L defines the wavelength, n is an energy level > 2 and R is the

R = 1.097 x 10^7 m^-1

From this one obtains (with appropriate algebra, change of units):

L = 3645 [n^2/ (n^2 – 4)]

which yields all wavelengths in Angstroms, unlike the original expression which yields wavelength in meters. (Bear in mind here the need to pay attention to the units, since 1 A = 10^-10 m )

Overlapping of lines must begin for that ‘n’ for which we have the condition:

L(n) – L(n + 1) = 1 A

Now, consider L (wavelength) to be a function of n as given above –such that we have:

3645 L^(-1) = 1 – 4n^(-2)

The next (and last) step in this solution hint, is that one makes use of differentials (after differentiating) such that:

-3645 L^-2 dL = 8n^-3 dn

From here, it is straightforward. Simply solve for n^3 on one side, with the derivative (-dn/dL) on the other. You then solve for n, and use the parameters of the problem which dictate:

dn = 1 and dL = -1

It's evident that this n is considerably greater than 1, so L must be very close to 3645A. Then:

n^3 ~ 8(3645), or n ~ 31 => (n + 1) = 1 A

That is, about 30 Balmer lines are visible.

2) The temperature of a gas is changed by a small amount. Derive an expression for the amount by which the

Solution:

The free electrons in an electron gas will produce a pressure given by:

P(e) = N(e) k T

Where k is the Boltzmann constant, k = 1.38 x 10^-23 J/K

Degrees of ionization are usually expressed by the partition function in logarithmic form:

log N(i+1)P(e)/ N(i) = -0.48 + log (2B(i+1)/B(i)) + 2.5 log T – 5040 I(i) / T

Where the second term on the right side bears the partition functions in terms of B.

If these partition functions can be taken as constants then the above equation shows that ionization degrees will remain unchanged IF:

2.5 log T – 5040 I/ T – log P(e) = 2.5 log (T + dT) – 5040 I/ (T + dT) – log(P(e) + dP(e)

One can then show - using appropriate substitutions - that the first equation can be reduced to the one above, using

log( 1 + dP(e)/ P(e)) = 2.5 log(1 + dT/ T) + 5040 I/ T [dT/ T + dT]

Now, use the approximation such that for SMALL values of x, one has:

log(1+ x) = 0.4343 (x – x^2/ 2 + …..)

And one derives the expression required, once dP(e)/ P(e), and dT/T are assumed very small, viz.

dP(e)/P(e) = dT/T {2.5 + 11,600I/T}

Therefore, an increase in temperature requires an increase in electron pressure in order to neutralize the ionization changes. The condition can't be satisfied for two or more elements because of the dependence upon the ionization potential I (though it can be approximately satisfied by elements having roughly the same I values)

http://brane-space.blogspot.com/2010/11/basic-problems-in-astrophysics-2.html

The caption notes that "

*an interesting problem is to obtain the distance modulus for a typical star in a globular cluster".*In fact, this is even more basic that it seems, once one assumes (on the basis of the diagram shown above) that all the cluster stars are essentially at the same distance (which they very nearly are given the spatial dimensions of the cluster are far far less than the distance to Earth). In this case, one has the apparent magnitude essentially equal to the absolute magnitude. Thus, a more interesting problem even, is to find the distance to such a cluster, if the apparent magnitude say, m = M = +18.Then solving the appropriate equation for the distance (in parsecs) yields:

d = 10^{(18 + 5)/ 5} pc = 10 ^4.6 pc = 39, 800 pc or about 40 kpc (kilo-parsecs) which is a very reasonable distance for a globular cluster.

Now, some more basic problems:

1) The lines of the Balmer series crowd close together as higher series members are considered. If each line were exactly one A (1 Angstrom) wide, how many Balmer lines would be individually visible without overlapping other lines?

Solution:

The Balmer lines are defined according to:

1/ L = R (1/ 4 – 1/n^2)

where L defines the wavelength, n is an energy level > 2 and R is the

*Rydberg constant*:R = 1.097 x 10^7 m^-1

From this one obtains (with appropriate algebra, change of units):

L = 3645 [n^2/ (n^2 – 4)]

which yields all wavelengths in Angstroms, unlike the original expression which yields wavelength in meters. (Bear in mind here the need to pay attention to the units, since 1 A = 10^-10 m )

Overlapping of lines must begin for that ‘n’ for which we have the condition:

L(n) – L(n + 1) = 1 A

Now, consider L (wavelength) to be a function of n as given above –such that we have:

3645 L^(-1) = 1 – 4n^(-2)

The next (and last) step in this solution hint, is that one makes use of differentials (after differentiating) such that:

-3645 L^-2 dL = 8n^-3 dn

From here, it is straightforward. Simply solve for n^3 on one side, with the derivative (-dn/dL) on the other. You then solve for n, and use the parameters of the problem which dictate:

dn = 1 and dL = -1

It's evident that this n is considerably greater than 1, so L must be very close to 3645A. Then:

n^3 ~ 8(3645), or n ~ 31 => (n + 1) = 1 A

That is, about 30 Balmer lines are visible.

2) The temperature of a gas is changed by a small amount. Derive an expression for the amount by which the

*electron pressure*must be changed in order that the relative abundances of the first two ionization stages of a given element be unchanged.Solution:

The free electrons in an electron gas will produce a pressure given by:

P(e) = N(e) k T

Where k is the Boltzmann constant, k = 1.38 x 10^-23 J/K

Degrees of ionization are usually expressed by the partition function in logarithmic form:

log N(i+1)P(e)/ N(i) = -0.48 + log (2B(i+1)/B(i)) + 2.5 log T – 5040 I(i) / T

Where the second term on the right side bears the partition functions in terms of B.

If these partition functions can be taken as constants then the above equation shows that ionization degrees will remain unchanged IF:

2.5 log T – 5040 I/ T – log P(e) = 2.5 log (T + dT) – 5040 I/ (T + dT) – log(P(e) + dP(e)

One can then show - using appropriate substitutions - that the first equation can be reduced to the one above, using

*the assumption of the constant partition functions*. (Note that the small d’s in the equation represent*deltas*- not exact differentials!)The preceding can be further reduced to:log( 1 + dP(e)/ P(e)) = 2.5 log(1 + dT/ T) + 5040 I/ T [dT/ T + dT]

Now, use the approximation such that for SMALL values of x, one has:

log(1+ x) = 0.4343 (x – x^2/ 2 + …..)

And one derives the expression required, once dP(e)/ P(e), and dT/T are assumed very small, viz.

dP(e)/P(e) = dT/T {2.5 + 11,600I/T}

Therefore, an increase in temperature requires an increase in electron pressure in order to neutralize the ionization changes. The condition can't be satisfied for two or more elements because of the dependence upon the ionization potential I (though it can be approximately satisfied by elements having roughly the same I values)

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