Answer
(a) $ X = \,^{228}\text{Th} $
(b) $ X = \,^{207}\text{Tl} $
(c) $ X = \,^{7}\text{Li} $
(d) $ X = \,^{60}\text{Ni} $
Work Step by Step
Let's analyze each process.
$$\color{blue}{\bf [a]}$$
$$ X \rightarrow \,^{224}\text{Ra} + \alpha $$
Here, $ X $ undergoes alpha decay, emitting an alpha particle ($ \alpha $ or $ \,^{4}\text{He} $), which has a mass number of 4 and an atomic number of 2.
We know that $ \,^{224}\text{Ra} $ has a mass number of 224 and an atomic number of 88, we can calculate the properties of $ X $:
- Mass number of $ X $: $ 224 + 4 = 228 $
- Atomic number of $ X $: $ 88 + 2 = 90 $
Thus, the element with atomic number 90 is thorium, so
$$ \boxed{X = \,^{228}\text{Th} }$$
$$\color{blue}{\bf [b]}$$
$$ X \rightarrow \,^{207}\text{Pb} + e^- + \bar{\nu} $$
This is a beta-minus decay process where $ X $ emits an electron ($ e^- $) and an antineutrino ($ \bar{\nu} $), which increases the atomic number by 1 while keeping the mass number unchanged.
Since $ \,^{207}\text{Pb} $ has a mass number of 207 and an atomic number of 82, the original isotope $ X $ had:
- Mass number: 207
- Atomic number: $ 82 - 1 = 81 $
Thus, the element with atomic number 81 is thallium, so
$$ \boxed{X = \,^{207}\text{Tl}} $$
$$\color{blue}{\bf [c]}$$
$$ \,^{7}\text{Be} + e^- \rightarrow X + \nu $$
This reaction represents electron capture, where $ \,^{7}\text{Be} $ captures an electron, resulting in a decrease in atomic number by 1.
We know that $ \,^{7}\text{Be} $ has a mass number of 7 and an atomic number of 4.
So after electron capture:
- Mass number remains the same: 7
- Atomic number decreases by 1: $ 4 - 1 = 3 $
Thus, the element with atomic number 3 is lithium, so
$$\boxed{ X = \,^{7}\text{Li} }$$
$$\color{blue}{\bf [d]}$$
$$ X \rightarrow \,^{60}\text{Ni} + \gamma $$
This reaction involves gamma decay, where the isotope $ X $ emits a gamma photon ($ \gamma $) without a change in mass or atomic numbers, indicating that $ X $ and $ \,^{60}\text{Ni} $ are the same isotope.
Since $ X $ and $ \,^{60}\text{Ni} $ (Nickel) have the same mass and atomic numbers, we conclude that
$$ X = \,^{60}\text{Ni} $$
