Provide the complete nuclear reaction for Holmium-144 undergoing positron emission:
¹⁴⁴₆₇Ho → ? + ⁰₊₁e

• A. ¹⁴⁴₆₇Ho → ¹⁴⁴₆₆Dy + ⁰₊₁e
• B. ¹⁴⁴₆₇Ho → ¹⁴⁷₆₆Dy + ⁰₊₁e
• C. ¹⁴⁴₆₇Ho → ¹⁴⁷₆₇Ho + ⁰₊₁e
• D. ¹⁴⁴₆₇Ho → ¹⁴⁸₆₇Ho + ⁰₊₁e

Answer: (A)