A radioactive nucleus (initial mass number Ao and atomic number Z) emits 3 α-particles and 2 positrons. The ratio of number of neutrons to that of protons in the final nucleus will be

Question

A radioactive nucleus (initial mass number Ao and atomic number Z) emits 3 α-particles and 2 positrons. The ratio of number of neutrons to that of protons in the final nucleus will be (A) (A-Z-8/Z-4) (B) (A-Z-4/Z-8) (C) (A-Z-12/Z-4) (D) (A-Z-4/Z-2)

Tardigrade Admin · Accepted Answer

In positive beta decay a proton is transformed into a neutron and a positron is emitted.

P^{+} ⟶ n^{0} + e^{+}

no. of neutrons initially was A - Z
no. of neutrons after decay (A - Z) - 3 x 2 (due to alpha particles) + 2 x 1 (due to positive beta decay)
The no. of proton will reduce by 8. [as 3 x 2 (due to alpha particles) + 2(due to positive beta decay)] Hence atomic number reduces by 8.

Q. A radioactive nucleus (initial mass number $A_{o}$ and atomic number Z) emits $3 α$ -particles and $2 p os i t ro n s$ . The ratio of number of neutrons to that of protons in the final nucleus will be

$\frac{A - Z - 8}{Z - 4}$

$\frac{A - Z - 4}{Z - 8}$

$\frac{A - Z - 12}{Z - 4}$

$\frac{A - Z - 4}{Z - 2}$

Q. A radioactive nucleus (initial mass number Ao​ and atomic number Z) emits 3α-particles and 2positrons. The ratio of number of neutrons to that of protons in the final nucleus will be

Z−4A−Z−8​

Z−8A−Z−4​

Z−4A−Z−12​

Z−2A−Z−4​

Q. A radioactive nucleus (initial mass number $A_{o}$ and atomic number Z) emits $3 α$ -particles and $2 p os i t ro n s$ . The ratio of number of neutrons to that of protons in the final nucleus will be

$\frac{A - Z - 8}{Z - 4}$

$\frac{A - Z - 4}{Z - 8}$

$\frac{A - Z - 12}{Z - 4}$

$\frac{A - Z - 4}{Z - 2}$