Question

The radius of first Bohr orbit of hydrogen atom is $0.529{A^ \circ }$ . Calculate the radii of $(i)$ the third orbit of $H{e^ + }$ ion and $(ii)$ the second orbit of $L{i^{2 + }}$ ion.

Answer 1

Hint : The Bohr model represents an atom as a small, positively charged nucleus surrounded by electrons. These electrons orbit the nucleus in spherical orbits, similar to the solar system in structure, but with electrostatic forces rather than gravity attracting them.

Complete Step By Step Answer:
The Bohr radius is a physical quantity equal to the most likely distance between the nucleus and the electron in the ground state of a hydrogen atom (non-relativistic and with an infinitely heavy proton). It was named after Niels Bohr because of its position in the Bohr atom model.
Let us consider the ${n^{th}}$ bohr orbit,
${r_n} = \dfrac{{{n^2}{h^2}}}{{4{\pi ^2}mZ{e^2}}}$
where ${r_n}$ represents the radius of orbit, $n$ represents the principle quantum number of the orbit and $Z$ represents the atomic number.
For hydrogen, we know that the atomic number $Z = 1$and since it is the first orbit $n = 1$
On substituting the above values, we get,
${r_1} = \dfrac{{{{(1)}^2}{h^2}}}{{4{\pi ^2}m(1){e^2}}} \\\ \Rightarrow {r_n} = \dfrac{{{h^2}}}{{4{\pi ^2}m{e^2}}} = 0.529{A^ \circ } \\\$
$(i)$ For $H{e^ + }$ ion, we know that the atomic number $Z = 2$and since it is the third orbit $n = 3$
On substituting the above values, we get,
${r_3}\left( {H{e^ + }} \right) = \dfrac{{{{(3)}^2}{h^2}}}{{4{\pi ^2}m(2){e^2}}} \\\ \Rightarrow {r_3}\left( {H{e^ + }} \right) = \dfrac{9}{2}\left[ {\dfrac{{{h^2}}}{{4{\pi ^2}m{e^2}}}} \right] \\\ = \dfrac{9}{2} \times 0.529 \\\ = 2.380{A^ \circ } \\\$

$(ii)$ For $L{i^{2 + }}$ ion, we know that the atomic number $Z = 3$and since it is the second orbit $n = 2$
On substituting the above values, we get,
${r_2}\left( {L{i^{2 + }}} \right) = \dfrac{{{{(2)}^2}{h^2}}}{{4{\pi ^2}m(3){e^2}}} \\\ \Rightarrow {r_2}\left( {L{i^{2 + }}} \right) = \dfrac{4}{3}\left[ {\dfrac{{{h^2}}}{{4{\pi ^2}m{e^2}}}} \right] \\\ = \dfrac{4}{3} \times 0.529 \\\ = 0.705{A^ \circ } \\\$

Therefore, the radii of $(i)$ the third orbit of $H{e^ + }$ ion is equal to $2.380{A^ \circ }$ and $(ii)$ the second orbit of $L{i^{2 + }}$ ion is equal to $0.705{A^ \circ }$.

Note :
1. The Bohr Model's Important Points are:
2. Electrons orbit the nucleus in fixed size and energy.
3. The orbit's energy is proportional to its size. In the smallest orbit, the lowest energy is contained.
4. When an electron travels from one orbit to another, it absorbs or emits energy.