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?

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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.

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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 #?

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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

Take a look at example #3:

https://www.chemteam.info/Radioactivity/Writing-Ne...

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