# The radius of the first bohr orbit (n=1) of hydrogen atom is 53.4 pm. The radius of bohr orbit having n=3 in $L{i^{2 + }}$will be :

A. 53.4 pm

B. 106.8 pm

C. 120.1 pm

D. 160.2 pm

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**Hint:**We know that neil bohr was the first to explain the general features of hydrogen atom structure and its spectrum. Bohr’s theory can be applied on the ions containing only one electron similar to that of hydrogen atom like $L{i^{2 + }}$, $B{e^{3 + }}$ and $H{e^ + }$, such species are also called hydrogen like species.

**Formula used:**For hydrogen like species, the radii expression from bohr’s theory is given as:

${r_n} = \dfrac{{{a_ \circ }({n^2})}}{Z}pm$

**Complete step-by-step solution:**

Let us understand the electron in the hydrogen atom can move around the nucleus in a circular path of fixed radius and energy. These paths are called orbits, stationary states or allowed energy states. These orbits are arranged concentrically around the nucleus.

This energy and radius does not change with time but only as one moves from lower stationary state to higher stationary state when the electron absorbs the required amount of energy and emits the energy to move from higher state to lower state. These stationary states or orbits are denoted by an integral number known as principal quantum number which varies from n=1, 2, 3, 4 etc.

The radii of these stationary states can be expressed as:

${r_n} = {n^2}{a_ \circ }$

where ${a_ \circ }$ is the radius of the first stationary state and is called Bohr radius.

In the question above we have been given with first bohr orbit with n=1 therefore we can calculate the value of ${a_ \circ }$by substituting the given values,

$

{r_1} = {(1)^2}{a_ \circ } \\

53.4 = {a_ \circ } \\

$

Similarly for hydrogen like species such as lithium ion the radii expression from bohr’s theory is given as

${r_n} = \dfrac{{{a_ \circ }({n^2})}}{Z}pm$

$\Rightarrow {r_n} = \dfrac{{53.4({n^2})}}{Z}pm$

Now we have been given n=3 and Z is the atomic number of lithium which is 3 therefore substituting we get,

${r_3} = \dfrac{{53.4({3^2})}}{3}pm = 160.2pm$

This energy and radius does not change with time but only as one moves from lower stationary state to higher stationary state when the electron absorbs the required amount of energy and emits the energy to move from higher state to lower state. These stationary states or orbits are denoted by an integral number known as principal quantum number which varies from n=1, 2, 3, 4 etc.

The radii of these stationary states can be expressed as:

${r_n} = {n^2}{a_ \circ }$

where ${a_ \circ }$ is the radius of the first stationary state and is called Bohr radius.

In the question above we have been given with first bohr orbit with n=1 therefore we can calculate the value of ${a_ \circ }$by substituting the given values,

$

{r_1} = {(1)^2}{a_ \circ } \\

53.4 = {a_ \circ } \\

$

Similarly for hydrogen like species such as lithium ion the radii expression from bohr’s theory is given as

${r_n} = \dfrac{{{a_ \circ }({n^2})}}{Z}pm$

$\Rightarrow {r_n} = \dfrac{{53.4({n^2})}}{Z}pm$

Now we have been given n=3 and Z is the atomic number of lithium which is 3 therefore substituting we get,

${r_3} = \dfrac{{53.4({3^2})}}{3}pm = 160.2pm$

**Hence, the correct option is D.****Note:**From the above we observe that the value of energy becomes more negative and that of radius becomes smaller with increase of atomic number Z, this means that the electron will be tightly bound to the nucleus.