An element A decays into an element C by a two-step process:
$A \to B + H e_{2}^{4}$ and $B \to C + {2 e}_{- 1}^{0}$ .Then
A and C are isotopes
A and C are isobars
B and C are isotopes
A and B are isobars

Question

An element A decays into an element C by a two-step process-Arightarrow B-He-2-4 and Brightarrow C- 2e - -1- 0 -ThenA and C are isotopesA and C are isobarsB and C are isotopesA and B are isobars

Toppr Admin · Accepted Answer

Conservation of mass no- and charge holds trueA-x2192-BZB-ZA-x2212-2MB-MA-x2212-4B-x2192-C-2eZC-ZB-2-ZAMC-MB-MA-x2212-4Different mass- but same atomic No-Therefore- A and C are isotopes-

An element A decays into an element C by a two-step process:A→B+He42 and B→C+2e0−1.ThenA and C are isotopesA and C are isobarsB and C are isotopesA and B are isobars

Conservation of mass no. and charge holds trueA→BZB=ZA−2MB=MA−4B→C+2eZC=ZB+2=ZAMC=MB=MA−4Different mass, but same atomic No.Therefore, A and C are isotopes.

An element A decays into an element C by a two-step process:
$A \to B + H e_{2}^{4}$ and $B \to C + {2 e}_{- 1}^{0}$ .Then
A and C are isotopes
A and C are isobars
B and C are isotopes
A and B are isobars

Conservation of mass no. and charge holds true
$A \to B$
$Z_{B} = Z_{A} - 2$
$M_{B} = M_{A} - 4$
$B \to C + 2 e$
$Z_{C} = Z_{B} + 2 = Z_{A}$
$M_{C} = M_{B} = M_{A} - 4$
Different mass, but same atomic No.
Therefore, A and C are isotopes.