In the decay series `._(92)U^(238)` to `._(82)Pb^(206)`, how many `alpha`-paritcles and how many `beta^(ɵ)`-particles are emitted?

In the decay series `._(92)U^(238)` to `._(82)Pb^(206)`, how many `alp

1.	In the decay series `._(92)U^(238)` to `._(82)Pb^(206)`, how many `alpha`-paritcles and how many `beta^(ɵ)`-particles are emitted?
Answer» The change in mass is 238 - 206 = 32 unit. It means that `32//4 = 8 alpha`-particles are emitted. With the emission of `8 alpha`-particles, the change in atomic number will be `8 xx 2 = 16`, i.e., the new element would have atomic number `92 - 16 = 76`. But the final product Pb has atomic number 82. It means there would have been an emission of 82 - 46 = 6 `beta`-particles.

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In the decay series `._(92)U^(238)` to `._(82)Pb^(206)`, how many `alpha`-paritcles and how many `beta^(ɵ)`-particles are emitted?

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