for A, Z, and X I'm typing it in the form A,Z,X (or X as a particle) since I obviously can't do sub or superscript...
1,1,H + 1,1,p --> 2.1,H + ?
I thought initially the missing particle at the end would be either 0,-1,e or 0,1,e (beta emission/decay) but apparently it's not. I am not sure how to figure this out then...
Are you trying to write this (but I'm putting the mass on top but on the other side)?
1H1 + 1p1 ==>2H1 + X
If so, the subscripts must add and the superscripts must add.
On th bottom w have 1+1 on the left which must add to 2 + 0 on the right. The top number is 1+1 on the left which must add to 1 + 1 on the right. Thus X must be oX1
I hope this comes out right because if I make any itsy bitsy slip, everything from then on gets printed wrong. Anyway, you identify X
I know they need to balance out on both sides. I am just confused because I am pretty sure it should be 0,+1, e ...e being X. Webassign is telling me this is wrong but if that's true I don't understand why and need that explained....
I don't think so. An electron is -1e^0. I think oX^1 must be a neutron, right.
neutron was my first answer...i am using +1 e^0 to be a positron though, not electron.
Oh i just realized how you typed the equation has the product flipped.
so it's 1,1,H + 1,1P --> 2,1H + ?
order is A,Z,X where A is mass and X is the particle or element
I don't think I flipped anything.
Your 2,1,H is my 2H^1
Your 1,1,H is my 1H^1
If your 2,1,H is the same as my 2H^1, then the emitted product is a neutron oN^1.
Hmm...I don't know. My webassign is telling me it's not a neutron. Thanks
If your equation is
1H^1 + 1p^1 ==> 1H^2, then X will be 1X^0 and that is a positron.
To determine the missing particle in a nuclear reaction, we need to balance the equation by conserving both mass number (A) and atomic number (Z) on both sides.
In the given reaction, we have:
1,1,H (proton) + 1,1,p (proton) -> 2.1,H (deuterium) + ?
Let's start balancing the equation.
On the left-hand side (LHS):
The atomic number (Z) is 1 (1 proton) + 1 (1 proton) = 2.
The mass number (A) is 1 (1 proton) + 1 (1 proton) = 2.
On the right-hand side (RHS):
The atomic number (Z) is 1 (1 proton).
The mass number (A) is 2 (2 protons + 1 neutron).
Since both Z and A need to be conserved in a nuclear reaction, the missing particle must have an atomic number (Z) of 1 and a mass number (A) of 1. This corresponds to the particle, 1,0,n (neutron).
Therefore, the completed equation would be:
1,1,H (proton) + 1,1,p (proton) -> 2.1,H (deuterium) + 1,0,n (neutron)