Scandium-44 ( $""_21^44Sc$ ) decays by emitting a positron. Which nuclide is the product of the decay?

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Answer 1 · 2017-05-26T21:17:48

The answer is $""_20^44Ca$

Here is a reference for Positron Emission .

Here is the equation:

$""_21^44Sc rarr ""_20^44Ca+e^+ + v_e$

Scandium-44 (""_21^44Sc""_21^44Sc) decays by emitting a positron. Which nuclide is the product of the decay?