Question

Complete the following reaction.

${}_{92}^{238}U{\to }_{90}^{234}Th+$................ + Energy

• A. ${}_2^{4}He$
• B. $4{}_{-1}^0\beta$
• C. $4{}_0^{1}Neutron$
• D. None of the above

Answer 1

The correct option is A ${}_2^{4}He$

The Mass number of particle emitted is $MassnumberofU-massnumberofTh=238-234=4$.

The Atomic number of particle emitted is $AtomicnumberofU-AtomicnumberofTh=92-90=2$.

So the emitted particle is $\alpha$ particle i.e., ${}_2^{4}He$