- #1

ant284

I have the following question

4Hydrogen->Helium+2electrons+2neutrino+gamma

In the sun, Hydrogen is converted into helium(see equation above) and that the reaction releases about 26.7MeV of energy. the sun radiates energy at the rate of 3.9*10^26W and has a mass about 1.99*10^30 kg of which 75% is hydrogen. Find out how long it will take the sun to convert 12% of its hydrogen into helium.

I'm abit blocked. anyhelp is appreciated

I think i have the right ideas.

For Four hydrogen into helium is 26.7Mev or 4.272*10^-12Joules.

now a hydrogen is about 1.008 or 1.0u or 1* (1.66*10^-27)kg.

so I do, (1.99*10^30) *.75=1.425*10^30kg and

i do answer * .12 to get those 12% of hydrogen to helium

1.42*10^30 * .12 = 1.79*10^29kg.

so i can do (1.79*10^29)/(1.66*10^-27) to get the number of Hydrogen

so i get 1.08*10^56 hydrogen.

since 4 Hydrogen into helium i divide by the total number of hydrogen to get

(1.08*10^56)/4=2.7*10^55

i multiply 2.7*10^55 by 26.7MeV to get the total energy output.

so i get

(2.7*10^55)*(4.272*10^-12)=1.15344*10^44

and since

watts=J/S, so i do 1/(3.9*10^26) * (1.15344*10^44) and i get

2.96*10^17 seconds.

I could get it into years by dividing the answer by (60*60*24*365) and i get about

9.4*10^9 years.

Any help is appreciated.