Question

Taking the Bohr radius as a₀ = 53 pm, the radius of Li⁺⁺ ion in its ground state, on the basis of Bohr’s model, will be about ______.

पर्याय

• 53 pm

• 27 pm

• 18 pm

• 13 pm

Answer 1

Taking the Bohr radius as a₀ = 53 pm, the radius of Li⁺⁺ ion in its ground state, on the basis of Bohr’s model, will be about 18 pm.

Explanation:

Bohr’s radius of orbit (for Hydrogen and H,-like atoms): For an electron around a stationary nucleus, the electrostatic force of attraction provides the necessary centripetal force.

i.e., `1/(4πε_0) ((Ze)e)/r^2 = (mv^2)/r`  ......(i)

Also `mvr = (nh)/(2pi)`  ......(ii)

From equations (i) and (ii), the radius of n^th orbit

`r_n = (n^2h^2)/(4pi^2 kZme^2) = (n^2h^2ε_0)/(pimZe^2) = 0.53 n^2/Z Å`  ......`(k = 1/(4πε_0))`

⇒ `r_n ∝ n^2/Z` or `r_n ∝ 1/Z`

`r_n = a_0 n^2/Z`, where a₀ = the Bhor radius = 53 pm

‘The atomic number (Z) of lithium is 3.

As `r_n = a_0 n^2/Z`,

Therefore, the radius of Li⁺⁺ ion in its ground state, on the basis of Bohr’s model. will be about `1/3` times to that of Bohr's radius.

Therefore, the radius of lithium ion is near r = `53/3` = 18 pm.