Question

Parto of the uranium decay series is shown
${}_{92}{U}^{238}{\to }_{90}T{h}^{234}{\to }_{91}P{a}^{234}\to$${}_{92}{U}^{234}{\to }_{90}T{h}^{230}{\to }_{88}R{a}^{226}$
How many pairs of isotopes are there in the above series :

• A. 1
• B. 2
• C. 3
• D. 0

Answer 1

Answer: (B)

2

Atoms of the element having same atomic number and different mass number are known as isotopes.

Thus ${}_{92}{U}^{238}$ is an isotope of ${}_{92}{U}^{234}$ whereas ${}_{90}T{h}^{234}$ is an isotope of ${}_{90}T{h}^{230}$.

Hence there are two pairs of isotopes.