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

1 Answer
Apr 28, 2017

Answer:

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

Explanation:

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"#