Of the following nuclides, the one most likely to be radioactive is

Question

Of the following nuclides, the one most likely to be radioactive is (A)  1531 P (B)  3066 Zn (C)  714 N (D)  614 C

Tardigrade Admin · Accepted Answer

614 C n=14-6=8 p=6 (n/p)=(8/6)=1.33 >1 Hence, 614 Chas (n/p) ratio greater than 1. It emits particles to have (n/p)=1. Thus, neutron changes to proton by emission of α-particles 614 C arrow 714 N + -10 e

Q. Of the following nuclides, the one most likely to be radioactive is

$_{15}^{31} P$

$_{30}^{66} Z n$

$_{7}^{14} N$

$_{6}^{14} C$

$_{6}^{14} C n = 14 - 6 = 8$
$p = 6$
$\frac{n}{p} = \frac{8}{6} = 1.33 > 1$
Hence, $_{6}^{14}$ Chas $\frac{n}{p}$ ratio greater than 1. It emits particles to have $\frac{n}{p} = 1$ .
Thus, neutron changes to proton by emission of $α$ -particles
$_{6}^{14} C \to_{7}^{14} N +_{- 1}^{0} e$

Q. Of the following nuclides, the one most likely to be radioactive is

1531​P

3066​Zn

714​N

614​C

614​Cn=14−6=8 p=6 pn​=68​=1.33>1 Hence, 614​ Chas pn​ ratio greater than 1. It emits particles to have pn​=1. Thus, neutron changes to proton by emission of α-particles 614​C→714​N+−10​e

$_{15}^{31} P$

$_{30}^{66} Z n$

$_{7}^{14} N$

$_{6}^{14} C$

$_{6}^{14} C n = 14 - 6 = 8$
$p = 6$
$\frac{n}{p} = \frac{8}{6} = 1.33 > 1$
Hence, $_{6}^{14}$ Chas $\frac{n}{p}$ ratio greater than 1. It emits particles to have $\frac{n}{p} = 1$ .
Thus, neutron changes to proton by emission of $α$ -particles
$_{6}^{14} C \to_{7}^{14} N +_{- 1}^{0} e$