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.
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