In the nuclear emission stated above: ${}_{82}^{290}X \xrightarrow{\alpha} Y \xrightarrow{e^+} Z \xrightarrow{

Question

In the nuclear emission stated above: ${}_{82}^{290}X \xrightarrow{\alpha} Y \xrightarrow{e^+} Z \xrightarrow{\beta^-} P \xrightarrow{e^-} Q$, the mass number and atomic number of the product Q respectively, are:

Accepted Answer

1. **Initial Nucleus ($X$):** Mass Number ($A$) = 290, Atomic Number ($Z$) = 82.
2. **Step 1: Alpha Decay ($\alpha$):** Emission of a Helium nucleus (${}_{2}^{4}He$).
   - $A' = 290 - 4 = 286$
   - $Z' = 82 - 2 = 80$
   - Nucleus $Y$ is ${}_{80}^{286}Y$.
3. **Step 2: Positron Emission ($e^+$):** Emission of a positron (${}_{+1}^{0}e$). $Z$ decreases by 1, $A$ remains unchanged.
   - $A'' = 286$
   - $Z'' = 80 - 1 = 79$
   - Nucleus $Z$ is ${}_{79}^{286}Z$.
4. **Step 3: Beta Minus Decay ($\beta^-$):** Emission of an electron (${}_{-1}^{0}e$). $Z$ increases by 1, $A$ remains unchanged.
   - $A''' = 286$
   - $Z''' = 79 + 1 = 80$
   - Nucleus $P$ is ${}_{80}^{286}P$.
5. **Step 4: Electron Emission ($e^-$):** Another $\beta^-$ decay. $Z$ increases by 1.
   - $A_{final} = 286$
   - $Z_{final} = 80 + 1 = 81$
   - Nucleus $Q$ is ${}_{81}^{286}Q$ .

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