A radioactive element ^24292X emits two -particles, one electron,… — Solution | JEE Physics

NVAEasyJEE 2023Radioactive Decay & Half-Life

A radioactive element $^{242}_{92}X$ emits two $\alpha$ -particles, one electron, and two positrons. The product nucleus is represented by $^{234}_{P}Y$ . The value of $P$ is _____.

Answer

Correct answer:87

Step-by-step solution

Standard Method

Given: The nucleus is $^{242}_{92}X$ . It emits two $\alpha$ -particles, one electron $\left(\beta^-\right)$ , and two positrons $\left(\beta^+\right)$ .

Find: The atomic number $P$ in $^{234}_{P}Y$ .

For radioactive decay:

Emission of an $\alpha$ -particle decreases mass number by $4$ and atomic number by $2$ .
Emission of an electron $\left(\beta^-\right)$ increases atomic number by $1$ .
Emission of a positron $\left(\beta^+\right)$ decreases atomic number by $1$ .

After two $\alpha$ -particles:

\text{Atomic number} = 92 - 2 \times 2 = 88, \quad \text{Mass number} = 242 - 2 \times 4 = 234

After one electron $\left(\beta^-\right)$ :

\text{Atomic number} = 88 + 1 = 89

After two positrons $\left(2\beta^+\right)$ :

\text{Atomic number} = 89 - 2 = 87

Therefore, the value of $P$ is $87$ .

Common mistakes

Treating $\beta^-$ emission as decreasing atomic number. This is wrong because an electron emission converts a neutron into a proton, so the atomic number increases by $1$ . Increase $Z$ by $1$ for each $\beta^-$ emission.
Forgetting that two $\alpha$ -particles are emitted. This is wrong because each $\alpha$ emission changes mass number by $-4$ and atomic number by $-2$ . Apply the change twice: mass number decreases by $8$ and atomic number by $4$ .
Changing the mass number during $\beta^-$ or $\beta^+$ emission. This is wrong because beta decay changes only the atomic number, not the mass number. Keep the mass number fixed at $234$ after the two $\alpha$ emissions.

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