Hydrogen to Helium conversion

In summary, the sun converts hydrogen into helium by fusion reactions. The energy released per reaction is based on the number of protons per kg of hydrogen and the mass of the hydrogen converted.
  • #1
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Hello, I'm having trouble with this question:

mass of proton= (1.674*10^-27)kg
mass of neutron= (1.675*10^-27)kg
mass of 4He= (6.648*10^-27)kg

How many Kg of Hydrogen does the sun convert into Helium per year?

Now my main problem is that I don't have the rate of conversion. I know that each 4 protons (H atoms) for one Helium atom, but at what rate? I multiplied mass of one proton by 4 to get (6.696*10^-27)kg but I'm not really sure how to convert this into Kg per year since I don't have a rate.

Any help is greatly appreciated,
thanks
 
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  • #2
Hello, I'm having trouble with this question:

mass of proton= (1.674*10^-27)kg
mass of neutron= (1.675*10^-27)kg
mass of 4He= (6.648*10^-27)kg

How many Kg of Hydrogen does the sun convert into Helium per year?

Now my main problem is that I don't have the rate of conversion. I know that each 4 protons (H atoms) for one Helium atom, but at what rate? I multiplied mass of one proton by 4 to get (6.696*10^-27)kg but I'm not really sure how to convert this into Kg per year since I don't have a rate.

Any help is greatly appreciated,
thanks
 
  • #3
b_o3 said:
Hello, I'm having trouble with this question:

mass of proton= (1.674*10^-27)kg
mass of neutron= (1.675*10^-27)kg
mass of 4He= (6.648*10^-27)kg

How many Kg of Hydrogen does the sun convert into Helium per year?

Now my main problem is that I don't have the rate of conversion. I know that each 4 protons (H atoms) for one Helium atom, but at what rate? I multiplied mass of one proton by 4 to get (6.696*10^-27)kg but I'm not really sure how to convert this into Kg per year since I don't have a rate.

Any help is greatly appreciated,
thanks

You have to look up the luminosity of the Sun to know how much energy is produced by the Sun in one year (L something of order 10^26 W so multiplying this by the number of seconds per year will give you the energy produced per year...Check the number, I am quoting from memory and I know it's not exactly 10^26)
 
  • #4
From a scientific video I remember the sun loses 4 million tons of mass every second.
 
  • #5
i have Luminosity= (3.9*10^26)joules/sec , I don't understand how to convert this to Kg of Hydrogen...
 
  • #6
I have the luminosity as (3.9*10^26) joules/sec. I am not sure how to convert this to Kg of Hydrogen though...
 
  • #7
I have the luminosity as (3.9*10^26) joules/sec. I am not sure how to convert this to Kg of Hydrogen though...
 
  • #8
b_o3 said:
I have the luminosity as (3.9*10^26) joules/sec. I am not sure how to convert this to Kg of Hydrogen though...

You have to find the energy released by each fusion reaction of hydrogen. How many Joules are produced by each reaction (hint: what is the change of mass? What equation does one use to then get the energy released?) Once you have the energy released by each reaction, take the total amount of energy produced by the Sun in one year and divide it by the energy produced per reaction. That will tell you how many reactions occur per year. Once you have that, multiply that number by the mass of hydrogen converted to helium in each reaction.
 
  • #9
haiha said:
From a scientific video I remember the sun loses 4 million tons of mass every second.

Yes, that's the correct number...where "ton" refers to metric tons, so 4 billion kg per second.
 
  • #10
well the conversion of hydrogen to He, does not exactly balance mass wise; you understand that part right. So you take the inital mass less the final mass to figure energy per fusion rxn--E=delta(mass)*c^2. The you just need to compute number of protons per Kg and divide by 4 then multiply by the energy liberated in a single rxn. This help, or is it the last part of the question?
 
  • #11
b_o3 said:
I have the luminosity as (3.9*10^26) joules/sec. I am not sure how to convert this to Kg of Hydrogen though...

You just apply the equation : E=mc^2
==> m=E/c^2.
With E=3.9e26 ; c=3.0e8 you get m=4.33e9kg/sec, quite similar to the value given by Haiha
 
  • #12
funny, i thought i hinted at same in a vanishing post several hours ago.
 
  • #13
I don't think you can just use hydrogen atoms, the fusion process involves deuterium and tritium.
 
  • #14
although these are indeed intermediaries, I think the overall stoichiometry of the process can be represented as above...Good point though.
 
  • #15
thanks guys
 

1. What is "Hydrogen to Helium conversion"?

"Hydrogen to Helium conversion" refers to the process of converting hydrogen atoms into helium atoms through nuclear fusion reactions.

2. How is this conversion achieved?

This conversion is achieved by combining two hydrogen atoms under extreme heat and pressure, causing them to fuse together and form a helium atom. This process releases a large amount of energy.

3. Why is this conversion important?

The conversion of hydrogen to helium is important because it is the main source of energy in stars, including our sun. It also has potential applications for clean and sustainable energy production on Earth.

4. Can this conversion be replicated on Earth?

While the process of hydrogen to helium conversion occurs naturally in stars, scientists are working on ways to replicate this process on Earth for energy production. However, it is still a complex and challenging process.

5. Are there any risks associated with this conversion?

There are currently no known risks associated with the conversion of hydrogen to helium. However, if this process is replicated on a large scale for energy production, safety precautions and proper waste management will need to be implemented.

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