Question

The ratio of the radii of first orbit of $H,H{e^ + }$ and $L{i^{2 + }}$ is

Answer 1

In this question we have to find the ratio of the radius of $H,H{e^ + }$ and $L{i^{2 + }}$ . So here we will use the formula that the radii of the first orbits of a hydrogen like ion is given by the formula: $r = \dfrac{{{n^2}}}{{Z{m_e}{k_e}{e^2}}}$ . We should know that in this formula that everything is constant except $Z$ , so we will assume the rest of the value and then solve it.

Complete answer:
Here we have $H,H{e^ + }$ and $L{i^{2 + }}$
We will use the formula
$r = \dfrac{{{n^2}}}{{Z{m_e}{k_e}{e^2}}}$ .
Let us assume
$K = \dfrac{{{n^2}}}{{{m_e}{k_e}{e^2}}}$ .
So we can write the formula as
$r = \dfrac{K}{Z}$ , where $Z$ is the number of orbits.
First we will solve for $H$ . We know that the orbit of $H$ is one, i.e.
$Z = 1$
So by putting the value in the formula, we have
$\Rightarrow r = \dfrac{K}{1}$ .
We will now calculate for $H{e^ + }$ . We know that the orbit of $H{e^ + }$ is two so we can write it as , i.e.
$\Rightarrow Z = 2$
So by putting the value in the formula, we have
$\Rightarrow r = \dfrac{K}{2}$ .
Now we will solve for $L{i^{2 + }}$ . We know that the orbit of $L{i^{2 + }}$ is three, i.e.
$\Rightarrow Z = 3$
So by putting the value in the formula, we have
$\Rightarrow r = \dfrac{K}{3}$ .
We will now write all of them together for their ratios:
$= \dfrac{K}{1}:\dfrac{K}{2}:\dfrac{K}{3}$ .
Since all the numerator are same, so we can cancel it out, it gives:
$\dfrac{1}{1}:\dfrac{1}{2}:\dfrac{1}{3}$
We will take the LCM and solve it:
$= \dfrac{{1 \times 6}}{{1 \times 6}}:\dfrac{{1 \times 3}}{{2 \times 3}}:\dfrac{{1 \times 2}}{{3 \times 2}}$
$= \dfrac{6}{6}:\dfrac{3}{6}:\dfrac{2}{6}$
Hence it gives us the ratio: $6:3:2$ .

Note:
We should know that we can also solve this question in an alternate way.
We should note that the radius of atom of Bohr’s equation is :
$r = \dfrac{{{a_0}{n^2}}}{Z}$ , where $r$ is the radius and $Z$ is the atomic number.
The value of ${a_0}$ is constant i.e. ${a_0} = 52.9$ .
Thus by putting these values again we get the same equation as above and we can solve it in a similar process.