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
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.)
4 answers
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.
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?
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)
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)