`._(90)Th^(232) to ._(82)Pb^(208)`. The number of `alpha and beta-"particles"` emitted during the above reaction is

Explanation

An `alpha`-particle is a helium nucleus with mass number 4 and atomic number 2. In an `alpha`-decay, the mass number of the disintegrating nucleus decreases by 4 and its atomic number decreases by 2.
A `beta^(-)` -particle is an electron with mass number 0 nad atomic number `-1`. In a `beta^(-)`-decay, the mass number of the disintegrating nucleus remains unchanged and its atomic number increases by 1.
Let `x alpha`-particles and `y beta^(-)`- particles be emitted in the disintegration of `""_(90)Th^(232)` into `""_(82)Pb^(200)`.
`""_(90)Th^(232) overset((xalpha + ybeta^(-)))to (""_(82)Pb^(200))`
`therefore 232 -4x-0(y)=200`
`therefore 4x=232 - 200 = 32`
`therefore x=8`
Also, `90-2x+1(y)=82`
`therefore 90 - 2(8)+y=82`
`therefore y=82 + 16-90=8`
`therefore 8 alpha`- particles and `8beta^(-)`-particles are emitted in the decay series of `Th^(232)` to `Pb^(200)`.