Question

Given the masses of various atomic particles ${m_p} = 1.0072\,u,\,\,\,\,{m_n} = 1.0087\,\,u,\,\,\,\,{m_e} = 0.000548\,\,u,\,\,\,\,{m_{\bar v}} = 0,\,\,\,\,{m_d} = 2.0141\,\,u,$ Where $p$ = proton, $n$ = neutron, $e$ = electron $\bar v = $ antineutrino and $d$ = deuteron. Which of the following process is allowed by momentum and energy conservation?

Accepted Answer

$n + p 	o d + \bar v$

Answer

$n + n 	o $ Deuterium atom (electron bound to the nucleus)

Answer

$p 	o n + {e^ + } + \bar v$

Answer

${e^ + } + {e^ - } 	o \gamma $

JEE PHYSICS MCQ: NUCLEI

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