Question

Question: Three atoms of element 111 (roentegenium) was produced in 1994 by bombarding ${}^{\mathbf{209}}\mathbf{Bi}$and ${}^{\mathbf{64}}\mathbf{Ni}$.

1. Write the balanced equation for the nuclear reaction. What other species is produced?
2. Write a balanced equation for the alpha decay process of this nuclide of roentegenium.

Short answer

1. ${}_{83}{}^{209}\mathrm{Bi}+{}_{28}{}^{64}\mathrm{Ni}\to {}_{111}{}^{272}\mathrm{Rg}+{}_0{}^1\mathrm{n}$
2. ${}_{111}{}^{272}\mathrm{Rg}\to {}_{109}{}^{286}\mathrm{Mt}+{}_2{}^4\mathrm{He}$
   

Step-by-step solution

Nuclear equation representing formation of roentegenium

When two lighter nuclei combine to form a heavier nuclide, the process is known as nuclear fission. The nuclear reaction taking place is represented by the nuclear equation below.

${}_{83}{}^{209}\mathrm{Bi}+{}_{28}{}^{64}\mathrm{Ni}\to {}_{111}{}^{272}\mathrm{Rg}+{}_0{}^1\mathrm{n}$

A neutron is emitted along with the formation of roentegenium.

Nuclear equation representing alpha decay

Alpha particle is a doubly ionized helium atom. When alpha decay takes place mass number of the nuclide decreases by 4 units while atomic number of the nuclide decreases by 2 units. The nuclear equation representing alpha decay by ${}_{111}{}^{272}\mathrm{Rg}$is:

${}_{111}{}^{272}\mathrm{Rg}\to {}_{109}{}^{286}\mathrm{Mt}+{}_2{}^4\mathrm{He}$