Chapter 21, Problem 21.124QA

Interpretation Introduction

To Find:

The supermassive element formed by the fission reaction.

Answer to Problem 21.124QA

Solution:

a) $Fe+ Bi \to Mt+n$

b) $Ni+ Bi \to Rg+ n$

c) $Ni+ Pb \to Ds+ n$

d) $Ne+ Bk \to Bh+4 n$

e) $Fe+ Pb \to Hs+ n$

Explanation of Solution

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:

$mass number \left(A\right)=number of neutrons +number of protons$

3) Given:

i) $Fe+ Bi \to ?+n$

ii) $Ni+ Bi \to ?+ n$

iii) $Ni+ Pb \to ?+ n$

iv) $Ne+ Bk \to ?+4 n$

v) $Fe+ Pb \to ?+ n$

4) Calculations:

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

a) $Fe+ Bi \to X+n$

The value of $A$ is calculated as

$58+209=A+1$

$A= \left(58+209\right)-1= \left(267\right)-1=266$

The value of $Z$ is calculated as

$26+83 =Z+0$

$Z= \left(26+83\right)-0= \left(26+83\right)=109$

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

The balanced nuclear reaction is

$Fe+ Bi \to Mt+n$

b) $Ni+ Bi \to X+ n$

The value of $A$ is calculated as

$64+209=A+1$

$A= \left(64+209\right)-1= \left(273\right)-1=272$

The value of $Z$ is calculated as

$28+83 =Z+0$

$Z= \left(28+83\right)-0= \left(28+83\right)=111$

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

The balanced nuclear reaction is

$Ni+ Bi \to Rg+ n$

c) $Ni+ Pb \to X+ n$

The value of $A$ is calculated as

$62+208=A+1$

$A= \left(62+208\right)-1= \left(270\right)-1=269$

The value of $Z$ is calculated as

$28+82 =Z+0$

$Z= \left(28+82\right)-0= \left(28+82\right)=110$

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

The balanced nuclear reaction is

$Ni+ Pb \to Ds+ n$

d) $Ne+ Bk \to X+4 n$

The value of $A$ is calculated as

$22+249=A+4\times 1$

$A= \left(22+249\right)-4= \left(271\right)-4=267$

The value of $Z$ is calculated as

$10+97 =Z+4\times 0$

$Z= \left(10+97\right)-0= \left(10+97\right)=107$

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

The balanced nuclear reaction is

$Ne+ Bk \to Bh+4 n$

e) $Fe+ Pb \to X+ n$

The value of $A$ is calculated as

$58+208=A+1$

$A= \left(58+208\right)-1= \left(266\right)-1=265$

The value of $Z$ is calculated as

$26+82 =Z+0$

$Z= \left(26+82\right)-0= \left(26+82\right)=108$

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 $Hs$.

The balanced nuclear reaction is

$Fe+ Pb \to Hs+ n$

Conclusion:

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