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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 ______.
Options
53 pm
27 pm
18 pm
13 pm
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Solution
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 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/(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.
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