Question

$^{}_{92}U^{238}$ on absorbing a neutron goes over to $^{}_{92}U^{239} .$ This nucleus emits an electron to go over to neptunium which on further emitting an electron goes over to plutonium. The resulting plutonium can be expressed as.

• A. $^{}_{94}Pb^{239}$
• B. $^{}_{92}Pb^{219}$
• C. $^{}_{93}Pb^{240}$
• D. $^{}_{92}Pb^{240}$

Answer 1

Answer: (A)

$^{}_{94}Pb^{239}$

Answer 2

The reaction can be shown as $^{}_{96}U^{238} + ^{}_{0}n^1 \, \, \longrightarrow \, \, ^{}_{96}U^{239} \longrightarrow$ $^{}_{93}Ne^{239} \, \longrightarrow \, \, ^{}_{94}Pb^{239} \, + \, \, ^{}_{-1}e^0$