When silver-107 is bombarded with a neutron, a different isotope of silver forms and then undergoes beta decay. What is the final product? (My Hints: Write two separate nuclear equations.)

I was thinking of something like this? (I'm not sure how to approach this problem):

108/47Ag ---> 107/47Ag + 1/0n then,
108/46Pd ---> 108/47Ag + 0/-1e

107/47 Ag + 1/0 n = 108/47 Ag

108/47 Ag ==> 108/48 Cd + 0/-1 e

Upper numbers must add up on both sides.
Lower numbers must add up on both sides.

Answer had to be expressed as just as an isotope so it is **108/48Cd

DrBob222 Can you explain how you got the 48 though? Where did the other proton come in when you just added to the mass?

The other proton, in somewhat more simplistic terms than may actually happen, comes from a neutron.

If we think of a neutron as consisting of a proton + an electron, the electron is ejected, we have an extra proton to add to those already present, and the mass number doesn't change since # protons + # neutrons is still 108 (1 less neutron and 1 more proton so sum is same as that initially)

To determine the final product when silver-107 is bombarded with a neutron and undergoes beta decay, we need to understand the process and the changes that occur during each step.

1. Bombardment of silver-107 with a neutron:
When a neutron is bombarded onto silver-107 (Ag-107), it undergoes neutron absorption, resulting in the formation of a new isotope of silver. The neutron is added to the nucleus, increasing the mass number of the atom.

Ag-107 + neutron → Ag-108

2. Beta decay of silver-108:
In beta decay, a neutron inside the nucleus is converted into a proton, emitting an electron and an antineutrino. This leads to an increase in the atomic number (Z) while the mass number (A) remains the same.

Ag-108 → Cd-108 + electron + antineutrino

So, the final product is cadmium-108 (Cd-108), which is formed after the beta decay of silver-108.

The resulting nuclear equations for the two steps are as follows:

1. Bombardment with a neutron:
Ag-107 + neutron → Ag-108

2. Beta Decay:
Ag-108 → Cd-108 + electron + antineutrino