2021 A/L Physics Past Paper · Question 28 (English Medium)

Answer: (3) 48 and 3

Given information

What we know:

A uranium nucleus takes in one slow neutron and splits into two smaller nuclei, $X$ and $Y$, and some neutrons.

$${}^{235}_{92}\text{U} + n \longrightarrow {}^{102}_{44}\text{X} + {}^{131}_{a}\text{Y} + bn$$

What each letter means:

$A$: mass number, the top number. It counts all the protons and neutrons in a nucleus.

$Z$: atomic number, the bottom number. It counts the protons.

$n$: a neutron. Its top number is $1$ and its bottom number is $0$, so it is ${}^{1}_{0}n$.

$a$: the atomic number of $Y$.

$b$: how many neutrons come out.

What the question asks

We want: the numbers $a$ and $b$.

The plan: count the top numbers on both sides to find $b$. Then count the bottom numbers on both sides to find $a$.

Relevant theory

Z is the atomic number or the number of protons in the nucleus.

A is the mass number.

The neutron is a neutral particle without any charge.

When a neutron is absorbed by a uranium 235 nucleus, it becomes unstable and then splits into two lighter nuclei which are more stable.

Required equations

$235 + 1 = 102 + 131 + b$

$92 + 0 = 44 + a + 0$

Applying it to this question

Think of a box of $236$ marbles that breaks open. The marbles roll into a few smaller piles. If you count every pile, you still have $236$ marbles. None are made and none are lost.

The nucleus is the same. Its protons and neutrons are only shared out into the new pieces. So the top numbers add up the same on both sides. The bottom numbers add up the same too.

First the top numbers. A neutron has top number $1$, so $b$ neutrons give $b \times 1 = b$.

$$235 + 1 = 102 + 131 + b$$

This says: all the protons and neutrons before the split equal all of them after.

$236$

$235 + 1$

$233$

$102 + 131$

$3$

$b = 236 - 233$

Now the bottom numbers. A neutron has no charge, so its bottom number is $0$. The neutrons add nothing here.

$$92 + 0 = 44 + a + 0$$

This says: the protons before the split equal the protons after.

$48$

$a = 92 - 44$

Reasoning and calculation

So $a = 48$ and $b = 3$. The reaction is:

$${}^{235}_{92}\text{U} + {}^{1}_{0}n \longrightarrow {}^{102}_{44}\text{X} + {}^{131}_{48}\text{Y} + 3\,{}^{1}_{0}n$$

It is option (3).

Checks

The unit: $a$ and $b$ are counts of particles, so they have no unit.

$au + 1 - ax - ay$

$zu - zx$

Does it make sense?

Top: $102 + 131 + 3 = 236$, and $235 + 1 = 236$. Bottom: $44 + 48 = 92$. Both sides match.

Uranium 235 often splits like this, setting a few neutrons free. Three is a sensible number.

Conclusion

$a = 48, \quad b = 3$

Top numbers: $235 + 1 = 102 + 131 + b$, so $b = 3$.

Bottom numbers: $92 + 0 = 44 + a$, so $a = 48$. The answer is (3).

In a nuclear reaction, add the top numbers on each side, then the bottom numbers on each side. Each pair must match.

Write a neutron as ${}^{1}_{0}n$ so you never forget its numbers, and never forget the neutron that comes in.

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