Write a balanced equation for:
1. beta decay of iodine-138
2. capture of an electron by cadmium-104
To write a balanced equation for the given processes, we need to understand the basic principles of these nuclear reactions.
1. Beta Decay of Iodine-138:
Beta decay is a nuclear process in which a beta particle (either an electron or a positron) is emitted from the nucleus. In the case of iodine-138, it undergoes beta-minus decay, meaning it emits an electron (beta particle). The general equation for beta decay is:
Parent Nucleus (Z,A) -> Daughter Nucleus (Z+1, A) + Beta Particle (0, -1)
Now we can substitute the values for iodine-138:
Iodine-138 (53,138) -> Xenon-138 (54,138) + Beta Particle (0, -1)
Therefore, the balanced equation for the beta decay of iodine-138 is:
^138_53I -> ^138_54Xe + ^0_-1β
2. Capture of an Electron by Cadmium-104:
In this process, cadmium-104 captures an electron and forms a new nucleus. This type of reaction is called electron capture. The general equation for electron capture can be written as:
Parent Nucleus (Z, A) + Electron (0, -1) -> Daughter Nucleus (Z-1, A)
Now we can substitute the values for cadmium-104:
Cadmium-104 (48,104) + Electron (0,-1) -> Silver-104 (47,104)
Hence, the balanced equation for the capture of an electron by cadmium-104 is:
^104_48Cd + ^0_-1e -> ^104_47Ag
1. The balanced equation for the beta decay of iodine-138 is:
^138_53I -> ^138_54Xe + 0_-1e
In this equation, iodine-138 decays into xenon-138 by emitting a beta particle (0_-1e) from its nucleus.
2. The capture of an electron by cadmium-104 can be represented by the following balanced equation:
^104_48Cd + 0_-1e -> ^104_47Ag
In this equation, cadmium-104 captures an electron to become silver-104.
It is difficult in this forum for us to write subscripts and superscripts. Here is what I'll do. The number to the left of the element symbol is the atomic number. The number to the right of the element symbol is the mass number. The way you do these problems is to make the subscripts and superscripts on EACH side add to the same value.In this case, you want the left numbers on both sides to add up and you want the right numbers to add up on both sides.
53I2138 ==> -1e0 + 54Y138
You know the beta particle is an electron (-1e0) so you write Y for the element that is left. Adding up subscripts on both sides you have 53 on the left and -1 on the right; therefore, the atomic number for Y MUST be 54 since -1 + x = 53 so x must be 54. Look on the periodic table and see what 54 is. That's Xe so replace Y with Xe. The superscript must be 138. The complete equation is
53I2138 --> -1e0 + 54Xe138
Now you do the last one.