Question

Atomic weight of boron is 10.81 and it has two isotopes ${}_{5}{{B}^{10}}$ and ${}_{5}{{B}^{11}}$. Then ratio of ${}_{5}{{B}^{10}}:{}_{5}{{B}^{11}}$ in nature would be
A. 19 : 81
B. 10 : 11
C. 15 : 16
D. 81 : 19

Answer 1

Hint: Use the formula for the average atomic weight of an atom that has two isotopes, $\text{average atomic weight = }\dfrac{\text{percentage of A }\\!\\!\times\\!\\!\text{ At}\text{.wt}\text{. of A + percentage of B }\\!\\!\times\\!\\!\text{ At}\text{.wt}\text{. of B}}{\text{100}}$.
Atomic weights of ${}_{5}{{B}^{10}}$ and ${}_{5}{{B}^{11}}$ is 10 and 11 respectively.

Formula used:
$\text{average atomic weight = }\dfrac{\text{percentage of A }\\!\\!\times\\!\\!\text{ At}\text{.wt}\text{. of A + percentage of B }\\!\\!\times\\!\\!\text{ At}\text{.wt}\text{. of B}}{\text{100}}$

Complete step-by-step solution -
The atomic number of boron is 5. It means that a boron atom consists of 5 electrons and 5 protons. The 5 protons are present inside the nucleus of the atoms along with neutrons.. The sum of the number of protons and the number of neutrons in the nucleus is called the mass number or atomic weight of the atom.
There exist atoms which have the same atomic number but different mass number. These types of atoms are called isotopes. The reason for different mass numbers for the same atoms is the different number of neutrons present in the nucleus.
Boron is an example. It has two isotopes ${}_{5}{{B}^{10}}$ and ${}_{5}{{B}^{11}}$. Each of the isotopes exist in nature in some abundance. We calculate the average atomic weight of all the boron atoms present in nature and generalise it as the atomic weight of boron.
The formula for the average weight of the atom is given as,
$\text{average atomic weight = }\dfrac{\text{percentage of A }\\!\\!\times\\!\\!\text{ At}\text{.wt}\text{. of A + percentage of B }\\!\\!\times\\!\\!\text{ At}\text{.wt}\text{. of B}}{\text{100}}$
Here, A and B are the isotopes of the atoms.
The question is asking to find the ratio of the percentage of the two isotopes of boron atoms.
Let the percentage of ${}_{5}{{B}^{10}}$ be X and the percentage of ${}_{5}{{B}^{11}}$ will be (100-X).
The atomic weight of ${}_{5}{{B}^{10}}$ be 10 and atomic weight of ${}_{5}{{B}^{10}}$ be 11.
Therefore, by the formula of atomic weight we get,
$10.81=\dfrac{X\times 10+(100-X)\times 11}{100}$
$\Rightarrow 1081=10X+1100-11X$
$\Rightarrow X=19$
Therefore, the percentage of ${}_{5}{{B}^{10}}$ is 19% and the percentage of ${}_{5}{{B}^{11}}$ is 81%.
Therefore, the ratio of ${}_{5}{{B}^{10}}:{}_{5}{{B}^{11}}$ is equal to 19 : 81.
Hence, the correct option is A.

Note: We know that particles with like charges repel each other due to electromagnetic forces. However, protons are present together inside the nucleus of the atoms.
A stronger force that the electromagnetic force called nuclear force holds the protons together.