Question

The ratio of the radius difference between ${4^{th}}$ and ${3^{rd}}$ orbit of H-atom and that of $L{i^{2 + }}$ ion is:
A. $1:1$
B. $3:1$
C. $3:4$
D. $9:1$

Answer 1

Hint: We can calculate the ratio of radius difference between 4th and 3rd orbital of H-atom and that of $L{i^{2 + }}$ ion by calculating the radius difference between 4th and 3rd orbit hydrogen atom and the radius difference between 4th and 3rd orbit of $L{i^{2 + }}$ ion. We can calculate the radius of any atom by using the formula,
${r_n} = \dfrac{{{n^2}}}{Z} \times 0.529$
Here n represents the principal quantum number of orbit
Z represents the atomic number

Complete step by step answer:
The value of n in the data is ${4^{th}}$ and ${3^{rd}}$ orbit.
We know the formula to calculate radius of any atom is,
${r_n} = \dfrac{{{n^2}}}{Z} \times 0.529$
For $L{i^{2 + }}$ ion, we know the atomic number is three. We can put the value of n as 4 and 3.
We have to substitute the value in the radius of $L{i^{2 + }}$ ion is,
Let us consider the value of n to be 4.
${r_n} = \dfrac{{{n^2}}}{Z} \times 0.529$
${r_4} = \dfrac{{{4^2}}}{3} \times 0.529$
${r_4} = \dfrac{{16}}{3} \times 0.529$
Let us take the value of n as 3.
${r_n} = \dfrac{{{n^2}}}{Z} \times 0.529$
$\Rightarrow {r_3} = \dfrac{{{3^2}}}{3} \times 0.529$
${r_3} = \dfrac{9}{3} \times 0.529$
Let us now calculate the difference of ${r_4} - {r_3}$ of $L{i^{2 + }}$ ion
$\Rightarrow {r_4} - {r_3} = 0.529\left( {\dfrac{{16}}{3} - \dfrac{9}{3}} \right)$
${r_4} - {r_3} = 0.529 \times \dfrac{7}{3}$
The radius difference of $L{i^{2 + }}$ ion is $0.529 \times \dfrac{7}{3}$.
For hydrogen atoms, we know the atomic number is one. We can put the value of n as 4 and 3.
We have to substitute the value in the radius of hydrogen atom is,
Let us consider the value of n to be 4.
${r_n} = \dfrac{{{n^2}}}{Z} \times 0.529$
$\Rightarrow {r_4} = \dfrac{{{4^2}}}{1} \times 0.529$
$\Rightarrow {r_4} = \dfrac{{16}}{1} \times 0.529$
Let us take the value of n as 3.
$\Rightarrow {r_n} = \dfrac{{{n^2}}}{Z} \times 0.529$
$\Rightarrow {r_3} = \dfrac{{{3^2}}}{1} \times 0.529$
${\Rightarrow r_3} = \dfrac{9}{1} \times 0.529$
Let us now calculate the difference of ${r_4} - {r_3}$ of hydrogen atom
${r_4} - {r_3} = 0.529\left( {\dfrac{{16}}{1} - \dfrac{9}{1}} \right)$
${r_4} - {r_3} = 0.529 \times 7$
The radius difference of the hydrogen atom is $0.529 \times 7$.
On equating the both equations, we get
$0.529 \times 7 = 0.529 \times \dfrac{7}{3}$
On taking the ratio, we will get
Ratio = $\dfrac{{\dfrac{7}{1}}}{{\dfrac{7}{3}}}$
Ratio = $3:1$
The ratio of the difference in the radius between ${4^{th}}$ and ${3^{rd}}$ orbit of H-atom and that of $\Rightarrow L{i^{2 + }}$ ion is $3:1$.

So, the correct answer is Option B.

Note: We know that the Bohr radius is a constant that is similar to the distance between the nucleus and the electron in an atom of hydrogen in its ground state. Hydrogen contains a single electron orbiting the nucleus and its smallest possible orbit that has lowest energy. The radius orbital of hydrogen is equal to Bohr radius. The value of Bohr radius is $5.291 \times {10^{ - 11}}m$.