Question

The uranium nucleus ${}_{92}{\text{U}}^{238}$ emits an $\alpha$ -particle and resulting nucleus emits one $\beta$ -particle. The atomic number and mass number of the final nucleus will be respectively

• A. $91,234$
• B. $90,234$
• C. $91,238$
• D. $92,234$

Answer 1

Answer: (A)

$91,234$

Release of a $\alpha$ will decrease the atomic number by 2 and mass number by 4, and $\beta$ particle will increase the mass number by 1 without change in mass number.

So final product will have a mass number of 234 and atomic number $=91$.