Question

The nuclear of which of following Pairs of nuclei are isotones (A) \({ }_{34} \mathrm{Se}^{74},{ }_{31} \mathrm{Ca}^{71}\) (B) \(_{42} \mathrm{Mo}^{92}, 40 \mathrm{Zr}^{92}\) (C) \({ }_{38} \mathrm{Sr}^{81}, 38 \mathrm{Sr}^{86}\) (D) \({ }_{20} \mathrm{Cd}^{40}, 16 \mathrm{~S}^{32}\)

Short answer

The pair of isotones is (A) \({ }_{34} \mathrm{Se}^{74},{ }_{31} \mathrm{Ca}^{71}\), as both nuclei have the same neutron number (N = 40).

Step-by-step solution

Calculate neutron numbers for each pair

For pair (A) \({ }_{34} \mathrm{Se}^{74},{ }_{31} \mathrm{Ca}^{71}\), neutron numbers are: N1 = \(A_{Se} - Z_{Se}\) = 74 - 34 = 40 N2 = \(A_{Ca} - Z_{Ca}\) = 71 - 31 = 40 For pair (B) \(_{42} \mathrm{Mo}^{92}, 40 \mathrm{Zr}^{92}\), neutron numbers are: N1 = \(A_{Mo} - Z_{Mo}\) = 92 - 42 = 50 N2 = \(A_{Zr} - Z_{Zr}\) = 92 - 40 = 52 For pair (C) \({ }_{38} \mathrm{Sr}^{81}, 38 \mathrm{Sr}^{86}\), neutron numbers are: N1 = \(A_{Sr81} - Z_{Sr}\) = 81 - 38 = 43 N2 = \(A_{Sr86} - Z_{Sr}\) = 86 - 38 = 48 For pair (D) \({ }_{20} \mathrm{Cd}^{40}, 16 \mathrm{~S}^{32}\), neutron numbers are: N1 = \(A_{Cd} - Z_{Cd}\) = 40 - 20 = 20 N2 = \(A_{S} - Z_{S}\) = 32 - 16 = 16

Find the pair of isotones

Now, comparing the neutron numbers for the given pairs: - Pair (A): N1 = 40 and N2 = 40, neutron numbers are equal, which indicates these nuclei are isotones. - Pair (B): N1 = 50 and N2 = 52, neutron numbers are not equal, which indicates these nuclei are not isotones. - Pair (C): N1 = 43 and N2 = 48, neutron numbers are not equal, which indicates these nuclei are not isotones. - Pair (D): N1 = 20 and N2 = 16, neutron numbers are not equal, which indicates these nuclei are not isotones. Only pair (A) has equal neutron numbers, so the correct answer is pair (A): \({ }_{34} \mathrm{Se}^{74},{ }_{31} \mathrm{Ca}^{71}\).