Question

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

Answer 1

Hint: 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.