Respuesta :
²⁵²Cf₉₈ + ¹⁰B₅ → ²⁵⁶Lr₁₀₃ + 6 ¹n₀
The new nuclide formed = Lawrencium - 256 = ²⁵⁶Lr₁₀₃
The new nuclide formed = Lawrencium - 256 = ²⁵⁶Lr₁₀₃
Answer: the nuclide is lawrencium-103
Using the isotopes notation it can be written as ²⁵⁶ ₁₀₃ Lr.
The number 256 is a superscript to the left of the symbol of the element (Lr) and the number 103 is a subscript to the left of the symbol
Explanation:
1) To determine the nuclide formed you have to rely in a mass and particles balance.
2) californium - 252 is the isotope has mass number 252, atomic number 98 (you obtain this information in a periodic table, remember that it is the number of protons).
The number of neutrons is determined from the formula
mass number = atomic number + number of neutrons ⇒ number of neutrons = mass number - atomic number
∴252 - 98 = 154 neutrons
3) boron - 10 nucleus has mass number 10, and atomic number 5 (from periodic table)
⇒ number of neutrons = 10 - 5 = 5.
4) Now the mass balance leads to:
mass number of californium-252 + mass number of boron-10 = unknown mass number + mass number of mass number of 6 neutrons
⇒ 252 + 10 = x + 6 ⇒ x = 252 + 10 - 6 = 256
5) Since neutrons do not change the atomic masses, number of protons, you can do this balance based on the atomic numbers (i.e. the same that number of protons):
atomic number of californium-252 + atomic number of boron - 10 = unknown atomic number
98 + 5 = A
A = 103
The element with atomic number 103 is lawrencium (Lr)
Calculate the number of neutrons: mass number - atomic number = 256 - 103 = 153.
Therefore, the new isotope is lawrencium (Lr) whith 153 neutrons.
Using the notation for isotopes the equation that show the reaction is:
²⁵² ₉₈ Cf + ¹⁰ ₅ B → ²⁵⁶ ₁₀₃ Lr + 6 ¹₀n