Question

Balanced nuclear equations for the following processes is:

Alpha emission of ${}^{162}Re$

• A. ${}_{75}^{162}Re{⟶}_{73}^{158}Ta{+}_2^{4}He$
• B. ${}_{74}^{162}Re{⟶}_{72}^{158}Ta{+}_2^{4}He$
• C. ${}_{76}^{162}Re{⟶}_{73}^{158}Ta{+}_3^{4}He$
• D. None of these

Answer 1

The correct option is A ${}_{75}^{162}Re{⟶}_{73}^{158}Ta{+}_2^{4}He$

The reaction involving alpha emission of ${}^{162}Re$ is as follows:

${}_{75}^{162}Re{⟶}_{73}^{158}Ta{+}_2^{4}He$

${}_2^{4}He=$ Alpha emission
Hence, option A is correct.