These are two questions and two complete answers.
Answers:
- Question #5.: alpha decay
Explanation:
A) Question #5.
The figure shows that the parent nuclide is [tex]^{222}_{86}Ra[/tex] and the daughter nuclide is [tex]^{218}_{84}Po[/tex]
A mass number balance and an atomic number balance show the features of the radiation emitted.
a) Mass number balance:
The mass number is the superscript to the left of the chemical symbol and is the amount of protons and neutrons in the nucleus of the atom.
- 222 = x + 218 ⇒ x = 222 - 218 = 4
Hence, the mass number of the unknown emitted particle is 4.
b) Atomic number balance.
The atomic number is the subscript to the left of the chemical symbol and is the number of protons in the nucleus of the atom.
- 86 = x + 2 ⇒ x = 86 - 2 = 84.
Hence, the atomic number (number of protons) of the unknown emitted particle is 2.
Conclusion: the emitted particle is a particle with 2 protons and 2 neutrons, which is a nucleus of helium, ⁴₂He, also called alpha particle, and the reaction is named alpha decay.
B) Question #6
Again, a mass number balance and an atomic number balance will tell which element is the missing product in the reaction.
The reaction is:
[tex]^{233}_{92}U+^1_0n{->}^{99}_{42}Mo+3^1_0n+?[/tex]
a) Atomic number balance:
- 92 + 0 = 42 + 3(0) + x ⇒ x = 92 - 42 = 50
Hence, the missing product has atomic number 50 which permits to tell that it is an atom of tin (Sn).
b) Mass number balance:
- 233 + 1 = 99 + 3(1) + x ⇒ x = 233 + 1 - 99 - 3 = 132
Which means that the missing product has a 132 neutrons and protons:
- 50 protons (determined above) and
- 132 - 50 = 82 neutrons.
The symbol of this atom is:
