16 give an example of a single gamma decay for your element...

16) Give an example of a single $\gamma$ decay for your element. Please write out the entire radioactive decay formula (both "reactant" and "product" sides.) [3 Marks] 17) Write out the decay series if your element were to undergo two $\alpha$ decays, a $\beta^-$ decay, and two $\beta^+$ decays. You need to write out each decay as its own radioactive decay series including the "reactant" and "products" each time for each decay. (So, in total, you should have 5 independent radioactive decay equations. Not a string of decays in a row.) [10 Marks]

Transcript

00:01 Hi, in this question we are having four parts.

00:03 In the first part of the question we are given that americium -241 isotope is bombarded with alpha particle as a result two neutrons are formed.

00:12 We are asked to write the balanced equation.

00:15 Here we can say helium particle is the alpha particle.

00:18 We can write 241 americium it is having the atomic number 95 and bombarded with alpha particle which is having the atomic number 2 and mass number 4.

00:30 Then there will be formation of x nuclei which will have the mass number b and atomic number as a and two neutrons are released here.

00:42 Here we need to find out the value of b and a.

00:45 We can equate the mass number on both side therefore here we can write 241 plus 4 equals to b plus 2 into 1.

00:57 On solving this we are getting the value of b equals to 243.

01:02 Let us calculate the a.

01:04 We can equate atomic number on both side we can write 95 plus 2 equals to a plus 2 into 0.

01:14 On solving we are getting the value of a equals to 97.

01:18 Hence the balanced equation will be written in this manner.

01:23 Here we are getting bercalium.

01:25 This is the answer of first part.

01:27 Now in second part we are given with cobalt 60 isotope and it is given that it undergoes beta emission followed by the gamma emission.

01:37 We can write the equation here as cobalt is having atomic number 27 and mass number is given as 60.

01:45 When it undergoes beta emission there is formation of x nuclei that will have atomic number a and mass number b and there is emission of beta particle in this manner and gamma particle in this manner.

02:00 We need to find out the value of a and b here.

02:04 We can equate the mass numbers and atomic numbers.

02:07 Let us find the b first.

02:08 We can write 60 equals to b plus 0.

02:13 It means b is equal to 60.

02:16 If we talk about a we can say 27 equals to a minus 1...

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