How do you write the nuclear equation for the beta decay of francium-223?

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${}_{87}^{233}{Fr} \to {}_{88}^{233}{Ra} + {}_{- 1}^{0}e + ν$

${}_{86}^{223}{Fr} \to {}_{88}^{233}{Ra} + {}_{+ 1}^{0}e + ν$

${}_{87}^{233}{Fr} \to {}_{90}^{223}{Ra} + {}_{- 1}^{0}e + ν$

${}_{87}^{233}{Fr} \to {}_{88}^{223}{Ra} + {}_{- 1}^{1}e + ν$

answer is A.

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Detailed Solution

We need to know the Francium-223 atomic number in order to solve this. We also need to understand how beta-decay reacts. The mass number of the element remains constant while the number of protons rises by one when a radioactive nucleus experiences beta decay.

The atomic number of francium is $87$ and the mass number of francium is $233$ . Thus, the element francium is denoted as ${}_{87}^{233}{Fr}$ .

Hence the nuclear equation for the beta decay of francium-223 is as follows:

${}_{87}^{233}{Fr} \to {}_{88}^{233}{Ra} + {}_{- 1}^{0}e + ν$