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 plutonium nucleus can be expressed as:

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

Answer 1

The correct option is A ${}_{94}P{u}^{239}$

$\text{Given}-{\ast }_{92}{U}^{238}\text{on absorbing neutron}{⟶}_{92}{U}^{239}$

$\text{Now the nucleus -}$

${}_{92}{U}^{239}\underset{–}{\text{emits an}{e}^{-}}{}_{93}N{p}^{239}$

$\begin{array}{c}\text{On further emitting an electron,}\\ {}_{93}N{p}^{239}\underset{–}{\text{emits an}{e}^{-}}{}_{94}P{u}^{239}\\ \text{Since, the emission of an}{e}^{-}\text{increases the atomic number}\\ \text{by (1) and no change in the atomic mass.}\\ \Rightarrow \text{opt (a) is correct.}\end{array}$