A radioactive element undergoes decay via the loss of two alpha particles to form a stable daughter isotope. Following the decay, what would the atomic number of this newly created stable isotope be?

Atomic number of the daughter isotope = (Atomic number of the parent isotope) - 4

Explanation:

An alpha particle has atomic number of 2 and a mass number of 4

Losing two alpha particles would lead to a reduction of 4 in atomic number and 8 in mass number for the radioactive isotope, to give the stable daughter isotope.

v=ir = 3x15 =45 v

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