A) Electron
B) Proton
C) Photon
D) Neutron
Correct Answer: D
Solution :
\[_{4}^{9}Be+_{2}^{4}He\xrightarrow[{}]{{}}\,_{6}^{12}C+_{z}^{A}X\] Comparing sum of mass numbers and atomic numbers on both sides, we get \[A=1,Z=0\] Hence, X represents \[_{0}^{1}n\] (neutron).You need to login to perform this action.
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