A) isomer
B) isotope
C) isobar
D) isomorph
Correct Answer: C
Solution :
\[_{1}{{H}^{3}}{{\xrightarrow{{}}}_{2}}H{{e}^{3}}{{+}_{-1}}{{e}^{0}}\]\[_{1}{{H}^{3}}\] and \[_{2}H{{e}^{3}}\] are isobars (having same mass no.) As, in the emission of \[\beta \]-particle, mass number does not change. Hence isobars are obtained. .
You need to login to perform this action.
You will be redirected in
3 sec
