Volume of nebula needed to form a star with same number of atoms as sun

[cattons3]

1. Question: In one region of a nebula the mean separation of atoms is 2.3x10^-3 m. What volume of the nebula contains enough atoms to make a star, similar to the sun, of radius 7.0x10^8 m within which the mean distance between atoms is 1.0x10^-10m?



2. Volume of sphere 4/3(pi(r^3 ))


3. Volume sun = 1.4x10^27 m^3. I thought that atoms in sun would be volume divided by separation distance and therefore volume of nebula that contains same number of atoms is atoms times separation distance (of nebula)? my answer was 3.22x10^34 m^3 answer in book is 1.8x10^49 m^3

I really hate these OCR A2 physics books because they don't work through any of the questions..
Help please!

 

[Ilon]

I agree i had the exact same question and objection to the books, i arrived at the same conclusion as you, did you ever resolve the question?

 

[cattons3]

I'm sorry that was a while ago now and I can't remember ...

 

[Ilon]

HAha fine, I'l go over it with my tutor tomorrow. Well done for putting your working so clearly, it helped.

 

[asteriatic]

I understand your frustration with the lack of explanation in some textbooks. Allow me to clarify the calculations for you.

Firstly, the volume of a sphere is indeed 4/3(pi(r^3)). However, in this case, we are not dealing with a perfect sphere as the nebula may have irregular shapes and densities. Therefore, we cannot simply use this formula to calculate the volume.

Instead, we can estimate the volume by assuming that the nebula is a cube with a side length of 2.3x10^-3 m (since the mean separation of atoms is given as 2.3x10^-3 m). This would give us a volume of (2.3x10^-3)^3 = 1.31x10^-8 m^3.

Next, we can calculate the number of atoms in this volume by dividing the volume of the sun (1.4x10^27 m^3) by the separation distance of the sun (1.0x10^-10 m). This gives us 1.4x10^37 atoms.

Therefore, the volume of the nebula needed to contain the same number of atoms as the sun would be 1.31x10^-8 m^3 x 1.4x10^37 atoms = 1.8x10^29 m^3.

I believe the answer in the book may have been a typo, as 1.8x10^49 m^3 is a much larger volume than the sun itself. I hope this helps clarify the calculations for you. Keep questioning and seeking understanding in your studies!