Question

Fill in the missing product in the following nuclear reactions.
a. 20881Tl ---> 20882Pb +?
b. 263106Sg ---> 42He + ?
c. ? ---> 0-1beta + 21483Bi
d. 8738Sr ---> 00gamma + ?

Answer 1

Hey @carlos1716 and welcome to \(\LARGE \sf \bbox[#40B9E9]{\color{white}{Open}}\bbox[#A8CE91]{\color{white}{Study!}}\)
You got any suggestions what the answers might be or want me to go over the theory and then we find the answer together?

Answer 2

We can go over the theory

Answer 3

Sure will.
In all nuclear reactions (and chemical reactions for that matter) there are some things that may just not be lost in the process:
1) matter
2) charge
3) energy

This will also be clear when we go over the questions.

In addtion to the above we have some decay types which we will use in order to determine the rest of the reaction equation.
The most common decay types are:
\(\alpha\) decay
\(\beta\) decay:
\(beta\) + (position emmision)
\(beta\) - (electron emmision)
\(\gamma\) decay.
Does any of the decay types say you something?

Answer 4

Or I like to make a correction: mass is not conserved. I'll be writing the theory done soon.

Answer 5

To go each of the mentioned decay types:

\(\Large \alpha - \sf decay\):
Is the decay of \(\large ^{4}_{2}\sf He\) cores.
We could writing the general reaction as:
\[\Large _{Z}^{A}\sf X \to _{\it Z \sf -2}^{\it A \sf -4}Y + _{2}^{4}He\]

\(\Large \beta - \sf decay\):
We can divide \(\beta\)-decay into two general reactions:
\(\large \beta^{-}-\sf decay\):
Is the decay of an electron, \(\large e^{-}\), and an antineutrino, \( \large \bar{v_{e}} \).
The general reaction could be written as follow:
\(\Large _{Z}^{A}\sf X \to _{\it Z \sf +1}^{\it A \sf }Y + ~ _{-1}^{0}\it e^{}\sf + \bar{v_{e}}\)

\(\large \beta^{+}-\sf decay\):
Is the decay of an positron (anti-electron), \(\large e^{+}\), and a neutrino \(\large v_{e}\).
The general reaction is:
\(\Large _{Z}^{A}\sf X \to _{\it Z \sf -1}^{\it A \sf }Y + ~ _{+1}^{0}\it e^{}\sf + v_{e}\)


\(\Large \gamma - \sf decay\):
Is the decay of a photon, \(\large \gamma\), with the energy \(\large hv\) (\( \large v\) is the frequency).
The general process is that the element goes from an excited state, \(\large \sf X^{*} \), into a more stable state, \(\large \sf X\), releasing the photon:
\(\Large _{Z}^{A}\sf X^{*} \to _{Z}^{A}\sf X + \gamma \)

You should be able to solve the problems from here. Otherwise just ask and I help you out.