Question

A problem relating to Nucleosynthesis

(a) Current scientific understanding is that Uranium is created in supernova explosions, with an initial U-235/U-238 ratio of 1.65 (that is, 1.65 U-235 atoms for every one U-238 atom). Determine the amount of time that has passed since the supernova event that formed Earth's uranium.* (10 pts) (b) From the luminosity of the Sun and the energy generated per Helium-4 nucleus created, determine the amount of Helium created per second, in kg, in the Sun's core. Compare this mass with an object of similar mass that you are familiar with (to put it in perspective). (5 pts)

Answer 1

Solution (a):

U-235/U-238 = 1.65............(1) (GIVEN)

Since their production, U-235 and U-238 have been decaying steadily with their respective half-lives. Therefore, if there were N0 nuclei present in the beginning then after time t, the number of nuclei remaining, N(t), will be given by

............(2)

2/1[/26 = ()N ON WA

where $T_{1/2}=$ half life of nuclei.

For U-235,
T1/2 = 7X108years
and for U-238, $T_{1/2}=4.5X10^{9}years$ ..........(3)

Also, let us consider that in the beginning there were of U-235. This implies, from (1),

for U-238..........(4)

From literature it is known that presently, U-235/U-238 ratio is

=> ...............(5). This is present value is also known from literature.

N|U-235 - 0.007 N|U-238

Using (3), (4) and (5) in (2), we get

NOU-235 24/7X108 N4\U-235 N U-238) present 1 No U-235 9t/4.5X10" beginning

...........(6)

Using (3), (4) and (5) in (6), we get,

=> .

0.00427 = 2-1.21X10-57

Taking natural logarithm on both sides, we get

. ANSWER

Solution (b):

Luminosity of Sun, ........(1) This is a known value from literature.

L = 3.82X1026J/S

Mass of ONE proton = ..........(2) This is a known value from literature.

=> Mass of 4 protons = .....(3)

Mass of ONE He nuclei = ...........(4) This is a known value from literature.

6.64X10-kg

Mass difference = Eq (3)-Eq(2)

$\Delta m=0.05X10^{-27} kg$..........(5)

This mass difference is converted to energy in Sun's core. This energy can be found using Einstein's equation

E = (Am)
............(6) where $c=3X10^{8}m/s$

From (5) and (6) we get,

$E=4.32X10^{-12}J$ for ONE reaction or for ONE He nucleus production...........(7)

Now using simple unitary method, we find

$4.32X10^{-12}J$ is produced by = $0.05X10^{-27}kg$ mass difference

=> $3.82X10^{26}J$ is produced by = $\frac{3.82X10^{26}X0.05X10^{-27}}{4.32X10^{-12}}=4.42X10^{9} kg$ per second is converted to energy in Sun's core...........(8)

Now, consider following

Number of reactions per second = L/E..........(9) From (1) and (7).

=> Number of reactions per second = $8.80X10^{37}$ reactions per second.........(10)

Now using simple logic, we can argue that if ONE reaction implies conversion of 4 protons to ONE He nucleus, then $8.80X10^{37}$ reactions will imply conversion of 4 times $8.80X10^{37}$ protons to nuclei.

Therefore, if mass of ONE He nucleus is
6.64X10-kg
THEN mass of $8.80X10^{37}$ He nuclei is

M = of He is produced per second in Sun's core. ANSWER

Solution (c):

[1] We will use the total luminosity of the sun (L). This includes both the electromagnetic $\left (3.82X10^{26} J/s \right )$ and neutrino $\left (0.089X10^{26} J/s \right )$ luminosities, given by

=> Energy, $E_{1}=3.939X10^{26}J$ ......(1)

[2] This energy will be divided by the energy released in ONE proton-proton chain reaction, which is given by

$E_{2}=26.7MeV=4.27X10^{-12}J$...(2)

[3] We will consider that this radiation will be spread over the surface of Earth. The shape of Earth's orbit can be considered to be a sphere for simplicity. Hence, the area will be given by;

$A_{1}=4 \pi r_{1}^{2}$ where $r_{1}=1.5X10^{11}m$ (radius of Earth's orbit around Sun)

=> $A_{1}=2.83X10^{23}m^{2}$ ..........(3)

[4] Area of human body, $A_{2}=1.7m^{2}$ .......(4)

Therefore, for a duration of one second, the number of neutrinos passing through human body is given by

$n=2\left (\frac{E_{1}}{A_{1}}/\frac{E_{2}}{A_{2}} \right )$...........(5). The factor 2 has been included because the proton-proton chain releases two neutrinos per reaction.

Using equations (1) to (4) in (5), we get

are crossing a body of area 1.7 sq. cm. ANSWER

Solution (d):

Cross sections basically gives the probability of interaction and is given by;

$\sigma =$ Number of neutrinos interacting with body (in 1 billion seconds)/Number of neutrinos passing through body (in 1 billion seconds) ....(1)

Now, Number of neutrinos passing through body in 1 s =

=> Number of neutrinos passing through body in = .........(2)

AND  $\sigma=10^{-14} barns$ (GIVEN)........(3)

Therefore, From (1), (2) and (3), we get

Number of neutrinos interacting with body in = ANSWER