MHT CET · Chemistry · Chemical Kinetics
In successive emission of \(\alpha\) -and \(\beta\) -particles, the number of \(\alpha\) -and \(\beta\) -particles should be emitted for the conversion of \({ }_{92} \mathrm{U}^{238}\) to \({ }_{82} \mathrm{~Pb}^{206}\) are
- A \(7 \alpha, 5 \beta\)
- B \(6 \alpha, 4 \beta\)
- C \(4 \alpha, 3 \beta\)
- D \(8 \alpha, 6 \beta\)
Answer & Solution
Correct Answer
(D) \(8 \alpha, 6 \beta\)
Step-by-step Solution
Detailed explanation
\({ }_{92} \mathrm{U}^{238} \longrightarrow{ }_{82} \mathrm{~Pb}^{206}+m_{2} \mathrm{He}^{4}+n_{-1} e^{0}\)
On comparing both sides
\(238 =206+4 m \)
\( \Rightarrow m =8 \)
\( 92 =82+2 m-n \)
\( 2 m-n =10 \)
\( \Rightarrow n =6\)
\(\therefore \alpha\) -particles emitted, \(m=8\) \(\beta\) -particles emitted, \(n=6\)
On comparing both sides
\(238 =206+4 m \)
\( \Rightarrow m =8 \)
\( 92 =82+2 m-n \)
\( 2 m-n =10 \)
\( \Rightarrow n =6\)
\(\therefore \alpha\) -particles emitted, \(m=8\) \(\beta\) -particles emitted, \(n=6\)
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