Question

An element E has two isotopes ${(}_{92}^{236}Eand{}_{92}^{238}E)$and the average atomic mass of element E is 237, then the total number of nucleons in its most abundant isotope is

• A.
  92
• B.
  235
• C.
  237
• D.
  238

Answer 1

The correct option is D

238
Atomic mass ${}_{92}^{236}E$ of isotope = 236 u

Atomic mass of ${}_{92}^{238}E$ isotope = 238 u

Let the abundance of ${}_{92}^{238}E$ isotope be $x\%$ and that of ${}_{92}^{236}E$ isotope be $(100–x)\%$.

Average atomic mass of the element $E=237u$

Applying the formula of average atomic mass, we get

$237=\frac{x\times 238+(100-x)\times 235}{100}23700=238x+23500-235x200=3xx=66.67\%100–x=33.33\%$

Therefore, the most abundant isotope of the element E is ${}_{92}^{236}E.$

$\because$ Mass number = Number of protons + Number of neutrons = Number of nucleons

$\therefore$ Total number of nucleons in the most abundant isotope = 238

Hence, the correct answer is option (d).