Chemistry Question?--- Nuclear Reaction?

the upper left number is the mass #.  it is the # of protons + # neutrons
the lower left # is the # of protons.  it's the same as the atomic number from a periodic table.

let's look at each of the particles given

.. 1
.. . .n
.. 0
.. . .# protons = 0
..  . # protons + # neutrons = 1
.. . .∴ # neutrons = 1 - 0 = 1
.. . .symbol "n" indicates a neutron

.. 10
.. . ..B
.. 5
.. . .# protons = 5 
.. . .# protons + # neutrons = 10
.. .. ∴ # neutrons = 10 - 5 = 5
.. . .symbol "B" is the symbol of Boron with atomic # = 5 = # protons

.. 4
.. . ..He
.. 2
.. . .# protons = 2
.. . .# protons + # neutrons = 4
.. .. ∴ # neutrons = 4 - 2 = 2
.. . .symbol "He" is the symbol of Helium with atomic # = 2 = # protons

so far so good right?

***********
they are 3 types of radioactive decay you need to know.
.. Alpha decay
.. beta - decay
.. beta + decay

in alpha decay, 
.. # neutrons decreases by 2
.. # protons decreases by 2.. . .. . . .. . (that lower left # decreases)
.. # neutrons + protons decreases by 4  (that upper left # decreases by 4)
.. a helium nucleus (that 4/2 He shown above is a helium nucleus) is emitted
.. OVERALL there is no net change in # of protons and # neutrons

in beta - decay 
.. # neutrons DECREASES by 1
.. # protons INCREASES by 1.. . .. (that lower left # increases by 1)
.. # neutrons + protons is constant (that upper left # is constant)
.. an electron (e-) + an antineutrino (ve+) are released
.. OVERALL.  There is a net change in # protons and # neutrons

in beta + decay
.. # neutrons INCREASES by 1
.. # protons DECREASES by 1.. . .. (that lower left # decreases by 1)
.. # neutrons + protons is constant (that upper left # is constant)
.. a positron (e+) + a neutrino (ve-) are released
.. OVERALL. There is a net change in # protons and # neutrons

usually,
.. we leave off neutrinos and sometimes electrons/positrons
.. on the left side of the rxn we assume # neutrons and # protons are conserved

example #1
.. 14.. .. . . 14
..  . .C ---> .. .N + 1 e + 1 antineutrino
.. ..6.. .. . . . 7 
.. . we can see that # protons increased by 1 while mass # remained constant
.. . this is an example of a beta - decay
.. . notice the OVERALL # of protons changed?  6 --> 7
.. ..and the overall # neutrons changed?  8 --> 7 (14-6 and 14-7)

example #2
.. 23.. .. . . 23
.. . .Mg ---> ..Na + 1 e(+) + 1 neutrino
.. 12.. .. . . .11
.. . we can see that # protons decreased by 1 while mass # remained constant
.. . this is an example of a beta + decay
.. . notice the OVERALL # of protons changed? 12 --> 11
.. ..and the overall # neutrons changed? 11 --> 12 (23-12 and 23-11)

example #3
.. 235.. .. .. 231.. .. .. . 4
.. . .U ---> ... .. Th + .. .. .He
.. 92.. .. . . . 90.. .. .. . .2
.. . notice in this case, the OVERALL # protons balances?  92 = 90+2?
.. ..notice in this case, the overall # neutrons balances?  143 = 141+2?
.. ..notice the He nucleus that is emitted?  4/2 He?
.. . .this is alpha decay.

***************
In general, you have to decide what kind of decay you're dealing with.. is it
.. (1) alpha?  is a 4/2 He nucleus emitted?
.. (2) beta-.. did we gain a proton while maintaining constant mass # ?
.. (3) beta+.. did we lose a proton while maintaining constant mass #?

**********
**********
your problem.
.. 1.. . . .10.. .. . . 11.. ... .. . 4.. .. . .?
.. . .n.. . . .B. --->.. . B ---->.. He + .. . ?
.. 0. . . . ..5.. .. .. ...5.. .. . . ..2.. .. . .?

we can see immediately that we have an alpha particle 4/2He produced so this is alpha emission and the overall net # protons must be constant and the overall net # neutrons must be constant.  so by balancing them

.. 1.. . . .10.. .. . . 11.. ... .. . 4.. .. . .7
.. . .n.. . . .B. --->.. . B ---->.. He + .. . ?
.. 0. . . . ..5.. .. .. ...5.. .. . . ..2.. .. . .3

right?  11 = 4 + 7 and 5 = 2 + 3
then we refer to a periodic table and see that the element with atomic # = 3 is lithium

.. 1.. . . .10.. .. . . 11.. ... .. . 4.. .. . .7
.. . .n.. . . .B. --->.. . B ---->.. He + .. LI..   <==== answer
.. 0. . . . ..5.. .. .. ...5.. .. . . ..2.. .. . .3...Show more

Take a look at example #3:

https://www.chemteam.info/Radioactivity/Writing-Neutron-Emission-Capture.html...Show more

Both the upper and the lower numbers must balance in the reaction. So, the mass number of the unknown product is:
1+10 = 4 + x
x = 7

The charge number of the unknown is:
0+5 = 2 + x
x = 3

The element with an atomic number of 3 is Lithium. Therefore, the product is:
7/3 Li...Show more