A radioactive nucleus (initial mass number A 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 A 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-4/Z-2) (B) (A-Z-8/Z-4) (C) (A-Z-4/Z-8) (D) (A-Z-12/Z-4)

Tardigrade Admin · Accepted Answer

When a radioactive nucleus emits an alpha particle, its mass number decreases by

4

while the atomic number decreases by

2

.
When a radioactive nucleus, emits a

β^{+}

particle (or positron

(e^{+})

) its mass number remains unchanged while the atomic number decreases by

1

.

_{z} X^{A} ⟶ 3 α_{z - 6} Y^{A - 12} ⟶ 2 e^{+}_{z - 8} W^{A - 12}

In the final nucleus,
Number of protons,

N_{p} = Z - 8

Number of neutrons,

N_{n} = A - 12 - (Z - 8)

= A - Z - 4

∴ \frac{N _{n}}{N _{p}} = \frac{A - Z - 4}{Z - 8}

Q. A radioactive nucleus (initial mass number $A$ 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

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

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

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

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

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

Z−2A−Z−4​

Z−4A−Z−8​

Z−8A−Z−4​

Z−4A−Z−12​

Q. A radioactive nucleus (initial mass number $A$ 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

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

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

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

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