Question

The alpha decay of iridium -174

Answer 1

¹⁷⁴₇₇Ir → ¹⁷⁰₇₅Re + ⁴₂He

Step-by-step explanation:

• When a radioisotope undergoes alpha decay its mass number decreases by four while its atomic number decreases by 2.

• The mass number of Iridium is 174
• Iridium has an atomic number 77
• Therefore, the decay equation will be;

¹⁷⁴₇₇Ir → ¹⁷⁰₇₅Re + ⁴₂He

• Iridium-174 undergoes alpha decay to form rhenium-170 which has a mass number of 170 and an atomic number 75.