Predict a likely mode of decay for each unstable nuclide. a. Sb-132

Video transcript

Hey everyone, we're asked to identify the expected mode of decay for Tillery um 137. Now if we go ahead and look at our periodic tables we find that Hillary um has an atomic number of 52. And we've learned that the stability of an isotope is dependent on its n oversee ratio. And since our C is in between the values of 40 and then our anniversary ratio must be equal to 1.50 in order to be considered stable. Now let's go ahead and calculate R. N. Oversee ratio. We can do so by taking our mass number and subtracting our atomic number then we can go ahead and divide it by our atomic number. Z Plugging in Our Values We Got 1 -52. All divided by 52. This will get us to a value of 1.63. As we can see our value is greater than 1.50. This means that we are above the Valley of Stability which is that 1.50 value. And since we are above the value of stability this means we have too many neutrons in our isotope. So our expected mode of decay is either going to be beta decay or neutron emission. And looking at our answer choices. It looks like neutron admission is not one of our choices. So our answer here is going to be be beta decay. So I hope this made sense and let us know if you have any questions