Consider the following radioactive decay process:
$A 84218 \overset{\alpha}{⟶}{A}_{1}\overset{{\beta}^{-}}{⟶}{A}_{2}\overset{\gamma}{⟶}{A}_{3}\overset{\alpha}{⟶}{A}_{4}\overset{{\beta}^{+}}{⟶}{A}_{5}\overset{\gamma}{⟶}{A}_{6}$
The mass number and the atomic number of ${A}_{6}$ are given by:

Question

Consider the following radioactive decay process:

$A 84218 \overset{\alpha}{⟶}{A}_{1}\overset{{\beta}^{-}}{⟶}{A}_{2}\overset{\gamma}{⟶}{A}_{3}\overset{\alpha}{⟶}{A}_{4}\overset{{\beta}^{+}}{⟶}{A}_{5}\overset{\gamma}{⟶}{A}_{6}$

The mass number and the atomic number of ${A}_{6}$ are given by:

Quantrex Academy · Accepted Answer

Correct answer: C. 210 and 80

When an alpha particle is emitted from nucleus the nucleus loses two proton and two neutrons. This means the atomic mass number decreases by 4 and the atomic number decreases by 2.

In beta decays, the mass number of the atoms remains the same, but the atomic number changes: the atomic number increases in the negative beta decay and decreases in the positive beta decay by $1$ , respectively.

In gamma decay, mass number and atomic number remains unchanged.

Step by step variation is given below:

$84218 A\overset{\alpha}{⟶} 82214 {A}_{1}\overset{{\beta}^{-}}{⟶} 83214 {A}_{2}\overset{\gamma}{⟶} 83214 {A}_{3}\overset{\alpha}{⟶} 81210 {A}_{4}\overset{{\beta}^{+}}{⟶} 80210 {A}_{5}\overset{\gamma}{⟶} 80210 {A}_{6}$

Consider the following radioactive decay process: A 84218 α ⟶ A _ 1...

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Correct answer

Step-by-step solution

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