Question

Find out the missing particle in the following nuclear reaction?
${}_{78}^{196}Pt{+}_0^{1}n{\to }_{78}^{197}Pt+(?)$

• A. Proton
• B. Neutron
• C. Electron
• D. Positron
• E. Gamma

Answer 1

The correct option is E Gamma

The given reaction : ${}_{78}^{196}Pt$ $+$ ${}_0^{1}n{\to }_{78}^{197}Pt$ $+$ ${}_Z^{A}X$

In a nuclear reaction, mass number and the atomic number are conserved.

$\therefore$ $196+1=197+A$ $⟹A=0$

Also, $78+0=78+Z$ $⟹Z=0$

Thus, the missing particle is gamma $\left(\gamma \right)$.