1) Calculate the wavelength of the gamma ray photon (in nm) which would be needed to balance the endothermic part of the triple –alpha fusion equation. (Recall here that 1 eV = 1.6 x 10 ^{-19} J)

__Solution:__

**g**

**) = hc/**

**l**

**= 6.7 keV**

**So:**

**6.7 keV**

**= (6.62 x 10**

^{-34}J-s) (3 x 10

^{8}m/s ) /

**l**

**l**= (6.62 x 10

^{-34}J-s) (3 x 10

^{8}m/s ) / 6.7 keV

Converting to consistent energy units, using 1 eV = 1.6 x 10

^{-19}J:

**l**= (1.98 x 10

^{-26}J-m)/ (1.07 x 10

^{-16}J) = 1.85 x 10

^{-10}m = 185 nm

2) Verify the second part of the triple-alpha fusion reaction, especially the Q-value. Account for any differences in energy released by reference to the gamma ray photon coming off and specifically, give the wavelength of this photon required to validate the Q.

__Solution:__

^{8}**Be +**

^{4}He**®**

^{12}C +**g**

**+ 7.4 MeV**

^{2}

The role (and value in energy) of the gamma ray photon can be obtained by using instead the value for carbon of 12.011 u and following the procedure shown in (1))

3) The luminosity or power of the Sun is measured to be L = 3.9 x 10^{26} watts. Use this to estimate the mass (in kilograms) of the Sun that is converted into energy every second. State any assumptions made and reasoning.

__Solution:__

^{26}J/s

**delivered per second then is:**

*energy*^{26}J = E = m c

^{2}so the mass converted to energy is:

^{2}= (3.9 x 10

^{26}J)/ (3 x 10

^{8}m/s )

^{2}

= 4.3 x 10^{9} kg

We have to assume the luminosity represents the actual macroscopic mass converted into energy and is a faithful reflection of all the fusion reactions underlying the conversion.

4) a) Show that u _{I } = A cos (ar) + B sin(ar)

^{2}u

_{I}/dr

^{2}+ a

^{2 }u

_{I}= 0.

__cosine solution__needs to be discarded.

__Solutions:__

u

_{I }= A cos (ar) + B sin(ar)

Then: du

_{I }/dr

_{ }= - a A sin (ar) + a B cos(ar)

d

^{2}u

_{I}/dr

^{2}= - a

^{2}A cos(ar) - a

^{2}B sin (ar) =

- a

^{2}[ A cos (ar) + B sin(ar)]

Or: d

^{2}u

_{I}/dr

^{2}= - a

^{2 }u

_{I}

Transposing:

d

^{2}u

_{I}/dr

^{2}+ a

^{2 }u

_{I}= 0.

**Since R = u/ r**

*the cosine solution must be discarded*lest we get an unwanted infinity. This leaves:_{I }= B sin(ar)

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