If radius of second Bohr orbit of the He⁺ion is 105.8 pm, what is the radius of third Bohr orbit of Li²⁺ion?

Explanation:

Bohr's Radius:

For an electron around a stationary nucleus, the electrostatic force of attraction provides the necessary centripetal force.

Mathematically the radius of an orbit can be given as:

\({r_n} = \frac{{0.53{n^2}}}{Z}{\rm{{\dot A}}}\)

Where:

Z = atomic number, n = orbit number

So r ∝ n²/Z

Calculation:

Given: 

For, Atom Helium (Z =2), Second Orbit n = 2, radius r_n = 105.8 pm

For Lithium Z = 3, Third Orbit n = 3, radius = ?

Mathematically the radius of an orbit can be given as:

\(\frac{r_1}{r_2} = \frac{{\frac{n_1^2}{Z_1}}}{\frac{n_2^2}{Z_2}}{\rm{{}}}\) where, r1 , n1 , and Z1 are the radius, orbit number, and the atomic number of helium ion and , r2 , n2 and Z2 are the radius, orbit number, and atomic number of Lithium-ion. 

 So,  \(\frac{r_1}{r_2} = \frac{{\frac{2^2}{2}}}{\frac{3^2}{3}}{\rm{{}}}\) 

→ r₁/r₂ = 2/3

→ r2 = 1.5 × 105.8 pm

→ r2 = 158.7 pm

Conclusion: