PRELIMINARY STAGE 2025
Our Lady of Grace SHS: 71 points
Benkum SHS: 50 points
Prampram SHS: 31 points
QUESTION
Boron-10 is bombarded with an alpha particle to produce nuclide Y and a neutron.
Identify Y.
ANSWER: Nitrogen-13
SOLUTION:
$A = (10 + 4) - 1$
$A = 14 - 1$
$A = 13$
$Z = (5 + 2) - 0$
$Z = 7$
$Z = 7 \implies \text{Nitrogen (N)}$
$\text{Y} = _{7}^{13}\text{N}$
---
PRACTICE QUESTIONS
1. QUESTION: ${}^{9}\text{Be}$ is bombarded with an alpha particle to produce nuclide Y and a neutron. Identify Y.
ANSWER: Carbon-12 (${}^{12}_{6}\text{C}$)
SOLUTION:
$A = (9 + 4) - (1)$
$A = 12$
$Z = (4 + 2) - (0)$
$Z = 6$
$Z = 6 \implies \text{Carbon (C)}$
${}^{9}_{4}\text{Be} + {}^{4}_{2}\text{He} \rightarrow {}^{12}_{6}\text{C} + {}^{1}_{0}\text{n}$
2. QUESTION: ${}^{7}\text{Li}$ is bombarded with an alpha particle to produce nuclide Y and a neutron. Identify Y.
ANSWER: Boron-10 (${}^{10}_{5}\text{B}$)
SOLUTION:
$A = (7 + 4) - (1)$
$A = 10$
$Z = (3 + 2) - (0)$
$Z = 5$
$Z = 5 \implies \text{Boron (B)}$
${}^{7}_{3}\text{Li} + {}^{4}_{2}\text{He} \rightarrow {}^{10}_{5}\text{B} + {}^{1}_{0}\text{n}$
3. QUESTION: ${}^{11}\text{B}$ is bombarded with an alpha particle to produce nuclide Y and a neutron. Identify Y.
ANSWER: Nitrogen-14 (${}^{14}_{7}\text{N}$)
SOLUTION:
$A = (11 + 4) - (1)$
$A = 14$
$Z = (5 + 2) - (0)$
$Z = 7$
$Z = 7 \implies \text{Nitrogen (N)}$
${}^{11}_{5}\text{B} + {}^{4}_{2}\text{He} \rightarrow {}^{14}_{7}\text{N} + {}^{1}_{0}\text{n}$
4. QUESTION: ${}^{13}\text{C}$ is bombarded with an alpha particle to produce nuclide Y and a neutron. Identify Y.
ANSWER: Oxygen-16 (${}^{16}_{8}\text{O}$)
SOLUTION:
$A = (13 + 4) - (1)$
$A = 16$
$Z = (6 + 2) - (0)$
$Z = 8$
$Z = 8 \implies \text{Oxygen (O)}$
${}^{13}_{6}\text{C} + {}^{4}_{2}\text{He} \rightarrow {}^{16}_{8}\text{O} + {}^{1}_{0}\text{n}$
5. QUESTION: ${}^{14}\text{N}$ is bombarded with an alpha particle to produce nuclide Y and a proton. Identify Y.
ANSWER: Oxygen-17 (${}^{17}_{8}\text{O}$)
SOLUTION:
$A = (14 + 4) - (1)$
$A = 17$
$Z = (7 + 2) - (1)$
$Z = 8$
$Z = 8 \implies \text{Oxygen (O)}$
${}^{14}_{7}\text{N} + {}^{4}_{2}\text{He} \rightarrow {}^{17}_{8}\text{O} + {}^{1}_{1}\text{p}$
6. QUESTION: ${}^{24}\text{Mg}$ is bombarded with an alpha particle to produce nuclide Y and a neutron. Identify Y.
ANSWER: Silicon-27 (${}^{27}_{14}\text{Si}$)
SOLUTION:
$A = (24 + 4) - (1)$
$A = 27$
$Z = (12 + 2) - (0)$
$Z = 14$
$Z = 14 \implies \text{Silicon (Si)}$
${}^{24}_{12}\text{Mg} + {}^{4}_{2}\text{He} \rightarrow {}^{27}_{14}\text{Si} + {}^{1}_{0}\text{n}$
7. QUESTION: ${}^{19}\text{F}$ is bombarded with an alpha particle to produce nuclide Y and a neutron. Identify Y.
ANSWER: Sodium-22 (${}^{22}_{11}\text{Na}$)
SOLUTION:
$A = (19 + 4) - (1)$
$A = 22$
$Z = (9 + 2) - (0)$
$Z = 11$
$Z = 11 \implies \text{Sodium (Na)}$
${}^{19}_{9}\text{F} + {}^{4}_{2}\text{He} \rightarrow {}^{22}_{11}\text{Na} + {}^{1}_{0}\text{n}$
8. QUESTION: ${}^{23}\text{Na}$ is bombarded with an alpha particle to produce nuclide Y and a proton. Identify Y.
ANSWER: Magnesium-26 (${}^{26}_{12}\text{Mg}$)
SOLUTION:
$A = (23 + 4) - (1)$
$A = 26$
$Z = (11 + 2) - (1)$
$Z = 12$
$Z = 12 \implies \text{Magnesium (Mg)}$
${}^{23}_{11}\text{Na} + {}^{4}_{2}\text{He} \rightarrow {}^{26}_{12}\text{Mg} + {}^{1}_{1}\text{p}$
9. QUESTION: ${}^{27}\text{Al}$ is bombarded with an alpha particle to produce nuclide Y and a proton. Identify Y.
ANSWER: Silicon-30 (${}^{30}_{14}\text{Si}$)
SOLUTION:
$A = (27 + 4) - (1)$
$A = 30$
$Z = (13 + 2) - (1)$
$Z = 14$
$Z = 14 \implies \text{Silicon (Si)}$
${}^{27}_{13}\text{Al} + {}^{4}_{2}\text{He} \rightarrow {}^{30}_{14}\text{Si} + {}^{1}_{1}\text{p}$