Question

A certain nucleus A (mass number 238 and atomic number 92) is radioactive and becomes a nucleus (mass number 234 and atomic number 90) by the loss of a particle. State the change in the form of a reaction.

• A. ${}_{91}^{238}A\to {}_{90}^{234}B+{}_2^{4}He$
• B. ${}_{92}^{238}A\to {}_{90}^{234}B+{}_2^{4}He$
• C. ${}_{92}^{238}A\to {}_{90}^{232}B+{}_2^{4}He$
• D. ${}_{92}^{235}A\to {}_{90}^{234}B+{}_2^{4}He$

Answer 1

The correct option is B ${}_{92}^{238}A\to {}_{90}^{234}B+{}_2^{4}He$

The Mass number of particle emitted is $MassnumberofA-massnumberofB=238-234=4$.

The Atomic number of particle emitted is $AtomicnumberofA-AtomicnumberofB=92-90=2$.

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

${}_{92}^{238}A\to {}_{90}^{234}B+{}_2^{4}He$