A radioactive element A decays into radioactive element C by the following processes in succession. A → B + 2 H e 4 ; B → C + 2 e − Then elements

Question

A radioactive element A decays into radioactive element C by the following processes in succession. A → B + 2 ​ H e 4 ​ ; B → C + 2 e − Then elements

TestHub · Accepted Answer

Let the atomic number and mass number of element $A$ be $Z$ and $A_{m}$ respectively. So, $A$ can be represented as $_{Z} A^{A_{m}}$ .

In the first decay process, $A$ emits an alpha particle ( $_{2} He^{4}$ ):
$_{Z} A^{A_{m}} \to_{Z - 2} B^{A_{m} - 4} +_{2} He^{4}$
Thus, element $B$ has atomic number $Z - 2$ and mass number $A_{m} - 4$ .

In the second decay process, $B$ emits two beta particles ( $2_{- 1} e^{0}$ ):
$_{Z - 2} B^{A_{m} - 4} \to_{Z} C^{A_{m} - 4} + 2_{- 1} e^{0}$
Thus, element $C$ has atomic number $Z$ and mass number $A_{m} - 4$ .

Comparing elements $A$ and $C$ , both have the same atomic number $Z$ but different mass numbers ( $A_{m}$ and $A_{m} - 4$ ). Elements with the same atomic number but different mass numbers are called isotopes.

Therefore, $A$ and $C$ are isotopes.

Answer: $A$ and $C$ are isotopes.

Physics - Modern Physics Question with Solution | TestHub

Options:

Doubts & Discussion