Question

An element ₉₂U²³⁸ losses one alpha particle, and then one β-particle and then again one β- particle to get the final nucleus _bQ^a.
(i) Write down the values of a and b.
(ii) State the relationship between the nuclei P and Q.

Answer 1

(i) Calculation:

$${\phantom{A}}_{\phantom{92}}^{\phantom{}}{\phantom{A}}_{\phantom{2}92}^{\phantom{2}}\mathrm{P}{\phantom{A}}^{238}\stackrel{\mathrm{\alpha }\text{-}\mathrm{particle}}{\to }{\phantom{A}}_{\phantom{90}}^{\phantom{}}{\phantom{A}}_{\phantom{2}90}^{\phantom{2}}\mathrm{X}{\phantom{A}}^{234}\stackrel{\mathrm{\beta }\text{-}\mathrm{particle}}{\to }{\phantom{A}}_{\phantom{91}}^{\phantom{}}{\phantom{A}}_{\phantom{2}91}^{\phantom{2}}\mathrm{Y}{\phantom{A}}^{234}\stackrel{\mathrm{\beta }\text{-}\mathrm{particle}}{\to }{\phantom{A}}_{\phantom{92}}^{\phantom{}}{\phantom{A}}_{\phantom{2}92}^{\phantom{2}}\mathrm{Q}{\phantom{A}}^{234}$$

$${\phantom{A}}_{\phantom{b}}^{\phantom{}}{\phantom{A}}_{\phantom{2}b}^{\phantom{2}}\mathrm{Q}{\phantom{A}}^a\mathrm{and}{\phantom{A}}_{\phantom{92}}^{\phantom{}}{\phantom{A}}_{\phantom{2}92}^{\phantom{2}}\mathrm{Q}{\phantom{A}}^{234}$$ being same. Hence a = 243 and b = 92

(ii) On examining P and Q, we find that their mass number are different, i.e., 238 and 234, but both have the same atomic number i.e., 92, hence P and Q isotopes of P.