Question

Consider the nuclear changes.

${}_{82}^{204}Pb\underset{\text{Positron emission}}{\overset{-{\beta }^{+}}{\to }}A\stackrel{-\alpha }{\to }B\underset{\text{Positron emission}}{\overset{-{\beta }^{+}}{\to }}C$

Which of the following statement(s) is/are correct?

• A. C is an isotone of ${}_{82}^{204}Pb$
• B. A is an isobar of ${}_{82}^{204}Pb$
  
• C. C is an isotope of A
• D. B is an isobar of C

Answer 1

Answer: (A), (B), (D)

B is an isobar of C

${\beta }^{+}$ is positron which is ${}_{+1}^{0}e$
$\alpha$ is helium nucleus which is ${}_2^{4}He$
The complete reaction sequence will be
${}_{82}^{204}Pb\underset{\text{Positron emission}}{\overset{-{\beta }^{+}}{\to }}{}_{81}^{204}Tl\stackrel{-\alpha }{\to }{}_{79}^{200}Au\underset{\text{Positron emission}}{\overset{-{\beta }^{+}}{\to }}{}_{78}^{200}Pt$
a) Two nuclei are said to be isotones if they have the same number of neutrons. So, C is an isotone of ${}_{82}^{204}Pb$.
b) If two or more nuclei have the same atomic weight but different atomic number then they are said to be isobars. Thus, A is an isobar of ${}_{82}^{204}Pb$.
c) If the nuclei have the same number of protons but different number of neutrons, then they are called isotopes. So, C is not an isotope of A.
d) B is an isobar of C.