How much energy is released in the fusion of 4 protons?

[najisalem2009]

Homework Statement

In a nuclear fusion reaciton 4 protons fuse together. The final result is a helium nucleus and a release of energy. The net reaction is 4p→⁴He+ energy. The mass of a helium nucleus is 6.64×10^-27kg. How much energy is released each decay?

Homework Equations

Mass lost X atmoic mass = ()
then E=MC²

The Attempt at a Solution

im having trouble figuring out how tog et the loss of mass I know the answer to the problem is 3.6*10^-12

The only way i know how to go about doing this is by
doing E = (6.64*10^-27kg) x (2.99x10⁸ m/s)²
that results in E=5.94 x 10-¹⁰

Can anyone point me in the right direction.

 

[Dick]

You want to use the mass difference between 4 protons and He as the m in E=mc^2. As a technical point that still won't give you the answer you stated because 4p->He+(energy) is a bit wrong. Did you look up the answer? That may not be the answer they are looking for since the gave you an oversimplified reaction equation.

 

[najisalem2009]

3.6x10^-12, was the answer given to me.

I tried working it out using the mass of 4 protons - the mass of 4 He but the answer is still not correct.

Mass Lost = (1.673x10^-27 X4)-(4.002602x4)
=-16.01amu

-16.01[STRIKE]amu[/STRIKE] x )6.64x10-27/1 [STRIKE]amu[/STRIKE])
=1.063X10^-25 kg

E=MC²
E= -1.063x10^-25 x (2.99x10⁸)²
e= -9.50 x 10^-9

 

[najisalem2009]

ive also tried working it out the way you said and it the answer still didnt come up to 3.6x10^-12.

Does anyone know what I am doing wrong?

 

[Redbelly98]

3.6x10^-12, was the answer given to me.

I tried working it out using the mass of 4 protons - the mass of 4 He ...

That is exactly the right idea , so there is probably a math error somewhere.

... but the answer is still not correct.

Mass Lost = (1.673x10^-27 X4)-(4.002602x4)
=-16.01amu

You're using kilograms for the proton mass, and amu for the helium. You shouldn't do that, the units have to be consistent for things to work out.

Try working in kilograms -- they told you the helium mass in kg, so just use that value for the helium.

 

[najisalem2009]

(6.68x10^-27)4-2.656x10^-26=-.1988x10^-16
E=-.1988x10^-16x(2.99x10⁸)²
E=-1.77x10^-9

No matter how i do it i can't seem to get the answer of 3.6x10^-12

 

[Redbelly98]

(6.68x10^-27)4-2.656x10^-26=-.1988x10^-16
E=-.1988x10^-16x(2.99x10⁸)²
E=-1.77x10^-9

No matter how i do it i can't seem to get the answer of 3.6x10^-12

It seems that you are not really reading what the problem statement says very carefully.

The mass of one proton is 1.674x10^-27 kg. We start with 4 protons, so use 4 times the mass of one proton. It is not 4 times 6.68x10^-27 kg.

We are told, in the problem statement, that the mass of a helium nucleus is 6.64x10^-27 kg. Question for you: how many helium nuclei are there? Hint: the statement that "the final result is a helium nucleus..." tells you how many helium nuclei there are. Hint #2: it is not 4 helium nuclei, it is fewer than that. So, do not multiply the helium mass times 4.

 

[najisalem2009]

Thanks a lot red belly.

(6.68x10^-27)-(6.64x10^-27)= 4x10^-29
E=4x10^-29x(2.99x10⁸)²
E=3.57x10^-12