`._86Rn^(222)` is converted into `._84Po^(218)` and `._93Np^(239)` is converted into `._94Pu^(239)`. Name the particles emitted in each case and write down the corresponding equations.

Explanation

Let the particle emitted in each case be represented as `._ZX^A`, . Therefore,
(1) `._86Rn^(222) to ._84Po^(218)+._ZX^A`
Using the law of conservation of mass number and charge number, we get
`222=218+A :. A=222-218=4`
`86=84+Z :. Z=86-84=2`
Now, `A=4 and Z=2` corresponds to an alpha particle `._2He^4`. Therefore, emitted particle is an alpha particle, and the equation is
`._86Rn^(222) to ._84Po^(218)+._2He^4`
(ii) `._93Np^(239) to ._94Pu^(239)+._ZX^A`
Using the law of conservation of mass number and charge number, we get
`239=239+A :. A=239-239=0`
`93=94+Z :. Z=93-94=-1`
Now, A=0 and Z=-1 correspond to electron `(._-1e^0)`. Therefore, emitted particle is a beta particle,
and the equation is `._93Np^(239) to ._94Pu^(239)+(._-1e^0)`