Chapter 19, Problem 2STP

Interpretation Introduction

Interpretation: Nuclear equation given has to be completed.

Concept introduction: Nuclear reactions are those where an unstable nuclide disintegrates in order to form a stable nuclide with emission of various radiations feasibly. It is called radioactivity of nucleus. The particles or rays emitted during the process are alpha, beta and gamma rays which cause a change in the identity of nucleus.

Beta particle is one that has unit negative charge and alpha particle is one that has 2 unit positive charges and is equivalent to helium atom ${}_2^4\text{He}$ .

Answer to Problem 2STP

Complete equation is ${}_{29}^{65}\text{Cu}+{}_6^{12}\text{C}\to \text{3}{}_0^1\text{n}{+}_{35}^{74}\text{Br}$ .Correct option is D.

Explanation of Solution

For reaction, ${}_{29}^{65}\text{Cu}+{}_6^{12}\text{C}\to \text{3}{}_0^1\text{n}+\text{?}$

Missing particle is ${}_{35}^{74}\text{Br}$ . It is so because 3 neutron particles are emitted on the reaction. In any nuclear equation, atomic number (Z) and mass remains (A) intact. Neutron has no Z. So the new element formed will have sum of 29 and 6. Due to loss of 3 neutron, new A comes out to 74.

Conclusion

After loss of 3 ${}_0^1\text{n}$ , A must 74 and Z must be 35, so option A,B and C are incorrect.

Complete equation is ${}_{29}^{65}\text{Cu}+{}_6^{12}\text{C}\to \text{3}{}_0^1\text{n}{+}_{35}^{74}\text{Br}$ .Correct option is D.