Balance these nuclear equations 104/47 Ag -> 0/-1e +

Answer 1

To balance the nuclear equation \( ^{104}{47} \text{Ag} \rightarrow ^{0}{-1} \text{e} + X \), you need to find the particle \( X \).

In this decay process, silver-104 (Ag) undergoes beta decay, where an electron (beta particle) is emitted. The decayed nucleus will typically be a different element or isotope.

1. Start by noting the mass numbers and atomic numbers:

   • The mass number of Ag (104) is on the left.
   • The atomic number of Ag (47) is also on the left.

2. The emitted beta particle \( ^{0}_{-1} \text{e} \) has:

   • A mass number of 0.
   • An atomic number of -1.

3. To find \( X \), we can set up the following equations:

   • Mass number: \( 104 = 0 + \text{mass number of } X \)
   • Atomic number: \( 47 = -1 + \text{atomic number of } X \)

From the mass number equation: \[ \text{mass number of } X = 104 - 0 = 104 \]

From the atomic number equation: \[ \text{atomic number of } X = 47 + 1 = 48 \]

4. The resulting nucleus with mass number 104 and atomic number 48 is \( ^{104}_{48} \text{Cd} \) (cadmium).

Thus, the balanced nuclear equation is: \[ ^{104}{47} \text{Ag} \rightarrow ^{0}{-1} \text{e} + ^{104}_{48} \text{Cd} \]