Question

After the emission of one $\alpha$ -particle followed by two $\beta$ -particles from ${}_{92}^{238}U$, the number of neutrons in the newly formed nucleus is:

• A. $140$
• B. $142$
• C. $144$
• D. $146$

Answer 1

Answer: (B)

$142$

${}_{92}^{238}U{\to }_2^{4}He{+}_{-1}^{0}e{+}_Z^{A}X$

$A=238-4+0=234$

$92=2-2+Z$

$Z=92$

No. of neutrons $=A-Z=234-92=142$