State what type of reaction this is

[amazingphysics2255]

Homework Statement

$^{235}_{~92}U+ ^{1}_{0}n \rightarrow \ ^{95}_{42}Mo+ ^{139}_{~57}La +2\, ^{1}_{0}n+\text X \, ^{~0}_{-1}e$
State what type of reaction this is, and solve for X

Relevant Equations

$^{235}_{~92}U+^{1}_{0}n \rightarrow ^{95}_{42}Mo +^{139}_{~57}La+2\, ^{1}_{0}n+\text X \, ^{~0}_{-1}e$

I believe It's fission, but am Unsure. As for X I will try solve later.

 

[kuruman]

It would be helpful if you corrected the LaTeX code to make your post legible. Please remember to use the "Preview" button before you post. Here is an example of how to write nuclides (just add one extra # on both sides), #^{235}_{92}U#. It's fission.

 

[amazingphysics2255]

$^{235}_{92}$U+$^{1}_{0}$$\rightarrow$$^{95}_{42}$Mo+$^{139}_{57}$+2$^{1}_{0}$n+X$^{0}_{-1}e$

sorry for that, how do I go about solving for X?

 

[collinsmark]

$^{235}_{92}$U+$^{1}_{0}$$\rightarrow$$^{95}_{42}$Mo+$^{139}_{57}$+2$^{1}_{0}$n+X$^{0}_{-1}e$

sorry for that, how do I go about solving for X?

(In the quoted text, you forgot to include the "La" [and "n"] symbol, although you did have that[them] in your original post.)

The reaction should balance not only the atomic mass numbers but also the atomic numbers on both sides of the reaction. In other words, the sum of the atomic mass numbers on the left hand side should all add up to the sum of the atomic mass numbers on the right hand side. Similarly, the atomic numbers should balance too.

When you see a large font number to the left of the particle, such as "2" in the $2 ^1_0 \rm{n}$ term, it means that there are two neutrons. That means when balancing the reaction, the atomic mass number and/or the atomic number of that particle needs to be counted twice.

You goal here is to find the large font number to the left of the beta particles. In other words, "how many beta particles are produced in this reaction?"

 

[amazingphysics2255]

$^{235}_{92}$U+$^{1}_{0}n$$\rightarrow$$^{95}_{42}$Mo+$^{139}_{57}La$+2$^{1}_{0}$n+X$^{0}_{-1}e$ 42+57=99 so does that mean X is going to be negative?

 

[Gaussian97]

Nop, $Z$ number is not conserved.

 

[kuruman]

42+57=99 so does that mean X is going to be negative?

This equation is partially OK, but it does not contain X which is what you are solving for. What must X be so that charge (the Z number) is conserved?

On Edit: Sorry, I meant to say, What must X be so that charge (not the Z number) is conserved?

 

[amazingphysics2255]

42+57=99 so 99-92=7 so is that negative 7? also @Gaussian97 said that Z is not conserved so Is this all wrong?

 

[Gaussian97]

I've said that $Z$ (the number of protons) is not conserved. If you interpret $Z$ as the electric charge, then it's different.

42+57=99 so 99-92=7 so is that negative 7? also @Gaussian97 said that Z is not conserved so Is this all wrong?

But no, -7 electrons make no sense.

 

[kuruman]

42+57=99 so 99-92=7 so is that negative 7? also @Gaussian97 said that Z is not conserved so Is this all wrong?

I omitted a very important "not" in post #7. I have edited the post and I am sorry for any confusion my omission might have caused.

 

[amazingphysics2255]

Okay, I'll try again. In the meantime are any sites that would be helpful for me to read?

 

[kuruman]

Try here.

 

[amazingphysics2255]

I can easily solve for Neutrons, and I can clearly see that the A is conserved 235+1=236. 95+139+2=236 conserved. Do I now need to do this for electrons/beta? if so 92+0=92. So for the charge to be conserved the end result of everything added up has to be 92 right? but 42+57+1=100 not 92 so that's why I was think X was going to be minus so we could get 92.

 

[kuruman]

There is negative charge and positive charge. Two negative charges added to one positive charge give one net negative charge. In these equations, it is the net charge, sum of positive and negative charge, that is conserved. In beta decay for example a neutron in the nucleus turns into a proton (still in the nucleus) and an electron is emitted plus a neutral antineutrino. The sum of charges on the left side (1 neutron) is zero and so is the sum of of charges on the right side, one positive proton and one negative beta or electron plus one zero for the antineutrino. The number of nucleons (protons + neutrons) is also conserved and is 1 on each side. Beta decay then can be expressed as $^A_Z X\rightarrow ~^A_{Z+1}Y+_{-1}e+\bar \nu$. Do you see how this works?

 

[collinsmark]

I can easily solve for Neutrons, and I can clearly see that the A is conserved 235+1=236. 95+139+2=236 conserved. Do I now need to do this for electrons/beta? if so 92+0=92. So for the charge to be conserved the end result of everything added up has to be 92 right? but 42+57+1=100 not 92 so that's why I was think X was going to be minus so we could get 92.

Where does that '1' come from (shown in red above)? I don't understand where that came from.

Let's take a step back. Here's the original reaction equation (I've taken the liberty to reformat the $\LaTeX$).

$$\rm{ ^{235}_{\ \ 92} U + \ ^1_0n \rightarrow \ ^{95}_{42}Mo + \ ^{139}_{\ \ 57}La + 2 ^1_0 n + X ^{\ \ \ 0}_{-1}e}$$

You've already balanced the atomic mass numbers in a previous post. That process of balancing goes like this:
235 + 1 = 95 + 139 + (2)(1) + ($x$)(0).
Of course, $x$ can be anything here because it's multiplied by 0. We can't solve for $x$ using this atomic mass number equation. So we'll have to proceed to balancing the atomic numbers to solve for $x$.

So now, do the same thing with the atomic numbers (in this notation, the beta particle's atomic number is represented by a "-1"). Write out the equation and solve for $x$.

(By the way, if you're wondering how I made the $\LaTeX$ look nice and readable, right-click the equation, go down to the "show math as" option, and click on "TeX commands.")

 

[amazingphysics2255]

92+0=42+57+(2)(0)+(x)(-1) make X=7 ?? But this conserves the number Z, not the charge, doesn't it?

 

[collinsmark]

92+0=42+57+(2)(0)+(x)(-1) make X=7 ?? But this conserves the number Z, not the charge, doesn't it?

Correct. There are 7 beta particles produced. And yes, the atomic number is conserved, because using the convention that your coursework uses, beta particles (free electrons) have an atomic number of -1.

[For future reference, hydrogen, deuterium, tritium and anti-beta particles (positrons) all have an atomic number of positive 1: $\ \rm{^1_1H, \ ^2_1H, \ ^3_1H, \ ^{\ \ \ 0}_{+1}e}$, but you can distinguish them by their unique atomic mass numbers if not the symbols.
(Sometimes deuterium and tritium are represented by $\ \rm{^2_1D}$ and $\ \rm{^3_1T}$ respectively, but I think your coursework uses $\ \rm{^2_1H}$ and $\ \rm{^3_1H}$.)]

 

[kuruman]

I was unaware that what I was taught to write as $e^+$ is now written as $^{\ \ \ 0}_{+1}e$. Is this a new convention for leptons?

 

[collinsmark]

I was unaware that what I was taught to write as $e^+$ is now written as $^{\ \ \ 0}_{+1}e$. Is this a new convention for leptons?

I'm not sure if it's new, but it's not uncommon. Although it might confuse chemists, it has many advantages for nuclear physics and nuclear engineering. And it does so without any loss of generality too. For nuclear reactions it's nice because it allows for easy balancing of nuclear reaction equations.

[Edit: for chemical reactions, the upper right hand corner is still available for use as the ionization number, so this convention doesn't "stomp" on the more established chemical reaction notations. E.g., you can denote an electron as $\ \rm{^{\ \ \ 0}_{-1}e^-}$ if you wish.]

 

[kuruman]

I'm not sure if it's new, but it's not uncommon. Although it might confuse chemists, it has many advantages for nuclear physics and nuclear engineering. And it does so without any loss of generality too. For nuclear reactions it's nice because it allows for easy balancing of nuclear reaction equations.

[Edit: for chemical reactions, the upper right hand corner is still available for use as the ionization number, so this convention doesn't "stomp" on the more established chemical reaction notations. E.g., you can denote an electron as $\ \rm{^{\ \ \ 0}_{-1}e^-}$ if you wish.]

That makes sense and is more consistent than the old way.