Answer: 1) The nuclide formed from the alpha decay of [tex]_{100}^{257}\textrm{Fm}[/tex] is [tex]_{98}^{253}\textrm{Cf}[/tex]
2) The nuclide formed from the beta-minus decay of [tex]_{98}^{253}\textrm{Cf}[/tex] is [tex]_{99}^{253}\textrm{Es}[/tex]
Explanation:
1) For alpha-decay
In this process, alpha particle is released.
Equation follows:
[tex]_{100}^{257}\textrm{Fm}\rightarrow _{98}^{253}\textrm{Cf}+_2^4\alpha[/tex]
The nuclide formed is [tex]_{98}^{253}\textrm{Cf}[/tex]
2) For beta-minus decay
In this process, a neutron is converted into a proton and an electron.
Equation follows:
[tex]_{98}^{253}\textrm{Cf}\rightarrow _{99}^{253}\textrm{Es}+_{-1}^0\beta[/tex]
The nuclide formed is [tex]_{99}^{253}\textrm{Es}[/tex]