MHT CET · Chemistry · Chemical Kinetics
An element \({ }_{x} A^{y}\) emits \(5 \alpha\) and \(4 \beta\) particles to give \({ }_{82} B^{207}\). The number of protons and neutrons in \(A\) are respectively
- A 88227
- B 88139
- C 82227
- D 84139
Answer & Solution
Correct Answer
(B) 88139
Step-by-step Solution
Detailed explanation
\({ }_{x} A^{y} \longrightarrow{ }_{82} B^{207}+5_{2} \mathrm{He}^{4}+4{ }_{-1} \beta^{0}\)
On comparing,
\(
\begin{aligned}
x &=82+5 \times 2+4(-1)=88 \\
y &=207+5 \times 4+4 \times 0 \\
&=227
\end{aligned}
\)
\(\operatorname{In}_{88} A^{227}\)
the number of protons \(=88\) the number of neutrons \(=227-88\)
On comparing,
\(
\begin{aligned}
x &=82+5 \times 2+4(-1)=88 \\
y &=207+5 \times 4+4 \times 0 \\
&=227
\end{aligned}
\)
\(\operatorname{In}_{88} A^{227}\)
the number of protons \(=88\) the number of neutrons \(=227-88\)
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