Question

Balanced nuclear equations for the following processes is:

Positron emission of ${}^{165}Ta$.

• A. ${}_{73}^{165}Ta{⟶}_{72}^{165}Hf{+}_1^{0}e$
• B. ${}_{74}^{165}Ta{⟶}_{73}^{165}Hf{+}_1^{0}e$
• C. ${}_{75}^{165}Ta{⟶}_{74}^{165}Hf{+}_1^{0}e$
• D. None of these

Answer 1

The correct option is A ${}_{73}^{165}Ta{⟶}_{72}^{165}Hf{+}_1^{0}e$

The reaction involving positron emission by ${}^{165}Ta$ is as follows:

${}_{73}^{165}Ta{⟶}_{72}^{165}Hf{+}_1^{0}e$

${}_1^{0}e=$ Positron emission

Hence, option A is correct.