What is the isotope remaining when uranium-235 decays in the following steps: $α, α, γ, β$ ?

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Answer 1 · 2017-04-28T21:06:00

The isotope remaining will be $""_89^227"Ac"$ .

First α decay

$""_92^235"U" → color(white)(l)_90^231"Th" +color(white)(l) _2^4"He"$

Second α decay

$""_90^231"Th" →color(white)(l) _88^227"Ra"^"*" + color(white)(l)_2^4"He"$

I presume that the radium isotope is in a metastable state, because it emits a γ ray in the next step

Gamma decay

$""_88^227"Ra"^"*" → color(white)(l)_88^227"Ra"+ γ$

β Decay

$""_88^227"Ra" → color(white)(l)_89^227"Ac" +color(white)(l) _text(-1)^0"e"$

What is the isotope remaining when uranium-235 decays in the following steps: α, α, γ, βα, α, γ, β?