TS EAMCET · Physics · Nuclear Physics
In the following nuclear reaction \(\mathrm{x}\) is \({ }_{13} \mathrm{Al}^{27}+{ }_2 \mathrm{He}^4 \rightarrow\) \({ }_0^{n^1+\mathrm{X}}\)
- A \({ }_{15} \mathrm{P}^{31}\)
- B \({ }_{14} \mathrm{Si}^{30}\)
- C \({ }_{15} \mathrm{P}^{30}\)
- D \({ }_{15} \mathrm{Si}^{31}\)
Answer & Solution
Correct Answer
(C) \({ }_{15} \mathrm{P}^{30}\)
Step-by-step Solution
Detailed explanation
\({ }_{13} \mathrm{Al}^{27}+{ }_2 \mathrm{He}^4 \longrightarrow{ }_0 \mathrm{n}^1+\mathrm{x}\) Mass no. remain conserve in nuclear reaction. \(A_x=27+4-1=30\) Atomic number remain conserve in nuclear reaction, \(\mathrm{Z}_{\mathrm{x}}=13+2-0=15\)…
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