Question

Which one of the following equations representing various nuclear processes can be balanced by filling in the blank with one alpha-particle?

• A. ${{}_{92}U}^{238}+{{}_{0}N}^1⟶{{}_{93}Np}^{239}+\_\_\_\_\_\_\_\_\_\_$
• B. ${{}_{7}N}^{14}+{{}_{0}n}^1⟶{{}_{6}C}^{14}+\_\_\_\_\_\_\_\_\_$
• C. ${{}_{38}Sr}^{90}⟶{{}_{39}Y}^{90}+\_\_\_\_\_\_\_\_$
• D. ${{}_{92}U}^{238}⟶{{}_{90}Th}^{234}+\_\_\_\_\_\_\_\_\_\_$

Answer 1

The correct option is D ${{}_{92}U}^{238}⟶{{}_{90}Th}^{234}+\_\_\_\_\_\_\_\_\_\_$

Alpha decay or $\alpha$ - decay is a type of radioactive decay in which an atomic nucleus emits an alpha particle (helium nucleus) and thereby transforms or 'decays' into a different atomic nucleus, with a mass number that is reduced by four and an atomic number that is reduced by two.

Therefore , in option D $U_{92}^{238}\to T{h}_{90}^{234}+H{e}_2^4$

The difference in atomic number is 2 . Hence, there is alpha decay.

Option D is correct .