Atoms in the Sun

Classical Mechanics Level 4

Calculate the approximate number of atoms in the sun. Give your answer as an order of magnitude .

This problem appeared in the 2014 UBC Physics Olympics (Problem 4)


The answer is 57.

Steven Zheng by Steven Zheng , 26, China

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4 solutions

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mass of sun= 19 × 1 0 30 19 \times 10^{30} kg and the sun is composed of hydrogen atoms whose weight is 1.67 × 1 0 − 27 1.67 \times 10^{-27} kg.........diving both the quantities to get the total no. of H atoms or total no. of atoms in sun we get 1.19 × 1 0 57 1.19 \times 10^{57} ....order is 57

6 Helpful 0 Interesting 0 Brilliant 0 Confused

At present we have: the mass of the Sun is:

Ms = 1.99 x 10^30 kg;

the mass of an atom of Hydrogen H, is mh = 1.67 x 10^-27 kg;

the mass of an atom oh Helium He is m(he) = 6.7 x 10^-27 kg;

the fraction in mass of Hydrogen is Ph = 0.71;

the helium one is P(he) = 0.28;

(approximately, I'm neglecting the other elements);

so: PhMs = mhNh; P(he)Ms = m(he)N(he);

and:

Nh = PhMs/mh = 0.71 x 1.99 x10^30/(1.67 x 10-27) =

Nh = 0.846 x 10^57;

N(he) = P(he)Ms/m(he) = 0.28 x1.99 x 10^30/(6.7 x10^-27)

N(he) = 0.083 x 10^57;

the number of atoms is:

N = Nh + N(he) = (0.846 + 0.083) x 10^57 = 0.968 x 10^57;

N ~ 0.929 x 10^57;

(we can observe that the fraction of Helium is very low in number of atoms);

The order of magnitude is: log(10^57) = 57

(but because I didn’t remember that the order is the exponent , I have not been able to answer correctly by the brilliant application...).

Antonio Fanari - 6 years, 9 months ago

@Steven Zheng Do you intend not to give the values like mass of the sun. You should give the values I believe

Ronak Agarwal - 6 years, 9 months ago

i like it its so nice and very interesting... i now learn wat the sun is nice one

John Steven Dela Serna - 6 years, 9 months ago

The question can be solved using "MOLE CONCEPT". Let the mass of sun be M. Mass of hydrogen=75% of M=0.75 M No. of hydrogen atoms=0.75 M 6.022 10^23 Mass of helium=25% of M No of helium atoms==0.0625 M 6.022*10^23

Therefore, total number of atoms=No of hydrogen atoms+no. of helium atoms

On putting the value of M as 2 10^30 and solving, we get- Total no. of atoms=9.785 10^56

As 9.785 is greater than 3.2, 10^56 is rounded off to 10^57

So, order of magnitude=57

I admit that the question requires a bit of general knowledge about the universe which should have been given in the question..... 1.mass of sun=2*10^30 kilograms 2.Sun comprises of 75% atomic hydrogen and 25% helium(which being a noble gas always exist in atomic state ) ..............without these figures, I don't think it is practical to solve the question.I might be wrong in my thinking.Please send me a better solution , which is sure to exist..........

3 Helpful 0 Interesting 0 Brilliant 0 Confused

I think that your reservations are well reasoned. :)

David Baker - 6 years, 9 months ago
John M.
Sep 5, 2014

If the solution is to divide the mass of the sun by the relative compositions of masses of hydrogen and helium atoms then the necessary quantities must be provided in the problem otherwise this becomes not a physics problem but a Jeopardy question.

The sun is composed only 75% of Hydrogen and 25% of Helium (approximately), according to Space.com .

If we do the calculations, we end up with a mass of approximately 1.6 × 1 0 57 1.6\times 10^{57} , taking the order of magnitude, 57 \boxed{57} .

3 Helpful 0 Interesting 0 Brilliant 0 Confused

ya, I absolutely agree with you on the fact that the figures required should have been given in the question........

Divyanshu Vadehra - 6 years, 9 months ago

It's a Fermi question.

Steven Zheng - 6 years, 9 months ago
David Baker
Sep 7, 2014

I actually did this accurately using %quantities of various elements present (in plasma form) in the sun [according to wikipedia] and got an answer (approx) of 9.55*10^56. They could have made it clearer that only hydrogen should have been accounted for. Other than that... great problem.

[Note if presuming 100% Hydrogen, you get 1.19*10^57]

0 Helpful 0 Interesting 0 Brilliant 0 Confused

I tried entering 1 followed by 57 zeros. and 1e+57 ... system didnt accept it! ..

Raghav Nagarajan - 6 years, 9 months ago

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