Question

Problem 5. [Einstein's E = mc²] Sun is producing energy by nuclear fusion reactions converting four hydrogen nuclei to form one nucleus of He, i.e. 4H + He. However, mass of helium nucleus is smaller than that of four hydrogens and this difference is called mass defect. If mass of the hydrogen nucleus is my and mass of the helium nucleus Mhe mass defect of one nuclear reaction Am is equal to Am = 4 *my - Me (1) Using Einstein's famous equation E = mc? we find that energy released in the single nuclear reaction corresponding to the mass defect Am is E = Am*c? (2) This is exactly where solar energy is coming from! Solar energy output in one second is, of course, solar luminosity Lsun. Thus in one second the total mass defect of the Sun is AM = LSun/c. (3) (a) Using equation (3), find the total mass AM that Sun loses in one second as mass defect. (b) Using equations (1) and (2) find the energy released in one nuclear reaction 4H'He. (c) Using the solar luminosity and your result from (b) find how many nuclear reac- tions are occurring in the solar core in one second as N = L Sun/E. (d) Given that in each reaction four hydrogens are transformed to helium, find the total mass of hydrogen that is spent in one second as Mtot = N*4* mh. Note: Even though we are talking about the nuclei, for my and Mue you can use the masses of hydogen and helium atoms, as for the purpose of this problem the difference in unimportant. Also, mass of other products of the reaction (positrons and neutrinos) is also negligible at the accuracy level of this problem - they are important only insomuch that they carry off energy, and that is already accounted for in the mass defect calculation.

Answer 1

1)

The luminosity of the Sun is 3.846 × 10^(26) watts

$\therefore \Delta m=\frac{3.846\times 10^{23}}{(3\times 10^{8})^{2}}=4.273\times 10^{6} Kg$

2)

$\Delta m=4m_{H}-m_{He}=4(Zm_{p}+Nm_{n})-(Z_{1m_{p}+N_{1}m_{n}})$

Z=no of protons in Hydrogen =1

N=no of neutrons in Hydrogen=0

Mp= mass of proton=1.00728u

Mn=mass of neutron=1.00867u

Z1=no of protons in Helium =2

N1=no of neutrons in helium=2

Am = 2.78 x 10-u= 2.78 x 10-3 x 1.6606 x 10-27 = 4.616 x 10-30 Kg

$E_{1}=4.1548\times 10^{-13} J$

3)

$N=\frac{L_{solar}}{E_{1}}=\frac{3.846\times 10^{26}W}{4.1548\times 10^{-13} J}=9.257\times 10^{38}s^{-1}$

4)

$M_{total}=N\times m_{H}=9.257\times 10^{38}s^{-1}\times 1.00728\times 1.6606\times 10^{-27}Kg=1.548\times 10^{13} kgs^{-1}$

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