Californium-252 is bombarded with a boron-10 nucleus to produce another nuclide and six neutrons. what nuclide forms?

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