The supermassive element formed by the fission reaction.

1) Concept:

To write the nuclear equations based on successful attempt to synthesize supermassive elements, we need to use the total mass of the reactant and product and also the total atomic weight on the reactant side and the product side. The nuclear reactions given in the examples are nuclear fission reactions. In this type of reactions, two nuclides fuse together to form a new nucleus with loss of neutrons. The mass number and the atomic number of the nuclei on the left hand side should be equal to the mass number and the atomic number of the fission products formed.

2) Formula:

mass number A=number of neutrons +number of protons

3) Given:

i) Fe2658+ Bi83209 → ?+n01

ii) Ni2864+ Bi83209 →?+ n01

iii) Ni2862+ Pb82208 → ?+ n01

iv) Ne1022+ Bk97249 → ?+4 n01

v) Fe2658+ Pb82208 →?+ n01

4) Calculations:

Let us assume that the nuclide formed is XZA, where A is the mass number of the nuclei and Z is the atomic number.

a) Fe2658+ Bi83209 → XZA+n01

The value of A is calculated as

58+209=A+1

A= 58+209-1= 267-1=266

The value of Z is calculated as

26+83 =Z+0

Z= 26+83-0= 26+83=109

Now, from the periodic table, we can see that the element with atomic number =109 is Meitnerium (Mt). Hence the super nuclide formed is Mt109266.

The balanced nuclear reaction is

Fe2658+ Bi83209 → Mt109266+n01

b) Ni2864+ Bi83209 → XZA+ n01

The value of A is calculated as

64+209=A+1

A= 64+209-1= 273-1=272

The value of Z is calculated as

28+83 =Z+0

Z= 28+83-0= 28+83=111

Now, from the periodic table, we can see that the element with atomic number =111 is Roentgenium (Rg). Hence the super nuclide formed is Rg111272.

The balanced nuclear reaction is

Ni2864+ Bi83209 → Rg111272+ n01

c) Ni2862+ Pb82208 → XZA+ n01

The value of A is calculated as

62+208=A+1

A= 62+208-1= 270-1=269

The value of Z is calculated as

28+82 =Z+0

Z= 28+82-0= 28+82=110

Now, from the periodic table, we can see that the element with atomic number =110 is Darmstadtium (Ds). Hence, the super nuclide formed is Ds110269.

The balanced nuclear reaction is

Ni2862+ Pb82208 → Ds110269+ n01

d) Ne1022+ Bk97249 → XZA+4 n01

The value of A is calculated as

22+249=A+4×1

A= 22+249-4= 271-4=267

The value of Z is calculated as

10+97 =Z+4×0

Z= 10+97-0= 10+97=107

Now, from the periodic table, we can see that the element with atomic number =107 is Bohrium (Bh). Hence the super nuclide formed is Bh107267.

The balanced nuclear reaction is

Ne1022+ Bk97249 → Bh107267+4 n01

e) Fe2658+ Pb82208 →XZA+ n01

The value of A is calculated as

58+208=A+1

A= 58+208-1= 266-1=265

The value of Z is calculated as

26+82 =Z+0

Z= 26+82-0= 26+82=108

Now, from the periodic table, we can see that the element with the atomic number =108 is Hassium (Hs). Hence the super nuclide formed is Hs108265.

The balanced nuclear reaction is

Fe2658+ Pb82208 →Hs108265+ n01

Conclusion:

From the atomic number of the reactant and product, we determined the unknown element in the balanced reaction.