Complete the following radioactive decay problem. 238\92u arrow 4/2he 234/90th arrow 234/91pa ( ? ) 4/2he 1/1p 1/0n 0/1e 0/-1e

Answer: 238/92 U → 4/2 He + 234/90 Th → 234/91 Pa + 0/-1e

• The first decay is considered as alpha decay 238/92 U → 4/2 He + 234/90 Th.
• The second decay of 234/90 Th is considered to be a beta decay that the change is in the atomic number from 90 to 91 so it losses 1 electron.
• So, the second decay will be: 234/90 Th → 234/91 Pa + 0/-1 e