Question

${}_{92}{U}^{238}$ absorbs a neutron. The product emits an electron. This product further emits an electron. The result is

• A. ${}_{94}P{u}^{239}$
• B. ${}_{90}P{u}^{239}$
• C. ${}_{93}P{u}^{237}$
• D. ${}_{94}P{u}^{237}$

Answer 1

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

${}_{92}{V}^{238}+n{\to }_{92}{A}^{239}$
${}_{92}{A}^{239}{\to }_{93}{B}^{239}+e$
${}_{92}{B}^{239}{\to }_{94}{C}^{239}+e$
Finding the element C from periodic table
${}_{94}P{u}^{239}$