6₁H¹ --> ₂He⁴ + 2₁H¹

4₁H¹ + 2₁H¹ --> ₂He⁴ + 2₁H¹

4₁H¹ --> ₂He⁴

 

Revered Members,

First equation is given in my text book to explain fusion reaction. Now i attempted the second and third reaction.

This is the fusion reaction happening in stellar energy. My doubt is , Can i cancel the 2₁H¹

 

found in LHS and RHS of second equation to arrive at the third equation?

 

 

Edited November 17, 201114 yr by logearav

6₁H¹ --> ₂He⁴ + 2₁H¹

4₁H¹ + 2₁H¹ --> ₂He⁴ + 2₁H¹

4₁H¹ --> ₂He⁴

 

Revered Members,

First equation is given in my text book to explain fusion reaction. Now i attempted the second and third reaction.

This is the fusion reaction happening in stellar energy. My doubt is , Can i cancel the 2₁H¹

 

found in LHS and RHS of second equation to arrive at the third equation?

 

 

 

 

Uhmm, depends on what you are trying to achieve I guess.

The net result is that you will get 1 Helium 4 from every 4 hydrogens, as the extra 2 hydrogens are available for future reactions (as you correctly surmised).

 

But that doesn't mean they aren't involved in the reaction (and in the extreme case of only having six hydrogen atoms, you would need them for the reaction).

 

 

The full reaction goes like this (D in this case is Deuterium, or Hydrogen-2):

[math] 2 _1H^1 \rightarrow 1 _1D^2[/math]

[math]1 _1D^2 + 1_1H^1 \rightarrow 1 _2 He^3 [/math]

This first part has to happen twice to give you two Helium-3 atoms (or just double all the numbers). Then:

[math] 2 _2He^3 \rightarrow _2He^4 + 2_1H^1 [/math]

 

Note that 3 hydrogens go into each Helium-3, and two are released when it becomes Helium-4.

If you only had 4 atoms you would not be able to do this reaction.