Can you write a balanced nuclear equation for the alpha decay of Ra-226?

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1 Answer
Ernest Z. · mrpauller.weebly.com
May 10, 2014

#""_88^226"Ra" → color(white)(l)_86^222"Rn" + color(white)(l)_2^4"He"#

Explanation:

An α-particle is a helium nucleus. It contains 2 protons and 2 neutrons, for a mass number of 4.

During α-decay, an atomic nucleus emits an alpha particle. It transforms (or decays) into an atom with an atomic number 2 less and a mass number 4 less.

Thus, radium-226 decays through α-particle emission to form radon-222 according to the equation:

#""_88^226"Ra" → color(white)(l)_86^222"Rn" +color(white)(l) _2^4"He"#

Note that the sum of the subscripts (atomic numbers or charges) is the same on each side of the equation.

Also, the sum of the superscripts (masses) is the same on each side of the equation.

Now try figuring out Am-241. Check your answer by watching the video below! Go to 3:35 if you want to skip the beta decay examples...

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