A) 5.0 MeV
B) 5.4 MeV
C) 5.6 MeV
D) 6.0 MeV
Correct Answer: B
Solution :
\[_{z}{{X}^{220}}\xrightarrow{\,}{{\,}_{2}}H{{e}^{4}}{{+}_{z-2}}{{Y}^{216}}+5.5\,\,meV\] \[|{{p}_{y}}|=|{{p}_{\alpha }}|\Rightarrow \,\sqrt{2{{m}_{y}}{{E}_{y}}}=\sqrt{2{{m}_{\alpha }}-{{E}_{\alpha }}}\] \[216{{E}_{y}}=4{{E}_{\alpha }}\] Also, \[{{E}_{\alpha }}+{{E}_{y}}=5.5\,\,MeV\] Solve to get K.E. of \[\alpha ={{E}_{\alpha }}=5.4\,MeV\]
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