KCET · Physics · Nuclear Physics
\({ }_{92} \mathrm{U}^{235}\) undergoes successive disintegrations with the end product of \({ }_{82} \mathrm{~Pb}^{203}\). The number of \(\alpha\) and \(\beta\) particles emitted are
- A \(\alpha=6, \beta=4\)
- B \(\alpha=6, \beta=0\)
- C \(\alpha=8, \beta=6\)
- D \(\alpha=3, \beta=3\)
Answer & Solution
Correct Answer
(C) \(\alpha=8, \beta=6\)
Step-by-step Solution
Detailed explanation
Let number of \(\alpha\) particles decayed be \(x\) and number of \(\beta\) particles decayed be \(y\).
Then equation for the decay is given by
\[
{ }_{92} \mathrm{U}^{235} \longrightarrow \mathrm{x} \alpha_{2}^{4}+\mathrm{y} \beta_{-1}^{0}+\mathrm{Pb}_{82}^{203}
\]
Equating the mass number on both sides
\[
235=4 x+203
\]
Equating atomic number on both sides
\[
92=2 \mathrm{x}-\mathrm{y}+82
\]
Solving Eqs. (i) and (ii), we get
\[
x=8, y=6
\]
\(\therefore 8 \alpha\) particles and \(6 \beta\) particles are emitted in disintegration.
Then equation for the decay is given by
\[
{ }_{92} \mathrm{U}^{235} \longrightarrow \mathrm{x} \alpha_{2}^{4}+\mathrm{y} \beta_{-1}^{0}+\mathrm{Pb}_{82}^{203}
\]
Equating the mass number on both sides
\[
235=4 x+203
\]
Equating atomic number on both sides
\[
92=2 \mathrm{x}-\mathrm{y}+82
\]
Solving Eqs. (i) and (ii), we get
\[
x=8, y=6
\]
\(\therefore 8 \alpha\) particles and \(6 \beta\) particles are emitted in disintegration.
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