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 ______.

Question

KollegeApply · Accepted Answer

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/(4$\pi$ε_0) ((Ze)e)/r^2 = (mv^2)/r ......(i) Also mvr = (nh)/(2pi)` ......(ii) From equations (i) and (ii), the radius of nth orbit `r_n = (n^2h^2)/(4pi^2 kZme^2) = (n^2h^2ε_0)/(pimZe^2) = 0.53 n^2/Z Å ......(k = 1/(4$\pi$ε_0)) $\Rightarrow$ 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 $\frac{1}{3}$ times to that of Bohr's radius. Therefore, the radius of lithium ion is near r = $\frac{53}{3}$ = 18 pm. [Solution diagram: 329184_solution_1.png]

AChoose Your Answer