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  • Taking the Bohr radius as a0 equal to 53pm, the radius of Li++ ion in its ground state, on the basis of Bohr’s model, will be about A. 53 pm B. 27 pm C. 18 pm D. 13 pm

1. Taking the Bohr radius as a0 = 53pm, the radius of Li++ ion in its ground state, on the basis of Bohr’s model, will be about
A. 53 pm
B. 27 pm
C. 18 pm
D. 13 pm

Answers (1)

The answer is the option (c).

 According to Bohr’s radius of the orbit, the necessary centripetal force is provided by the electrostatic force of attraction for hydrogen and H2 like atoms. So, the correct option is (c)18 pm as -

 i.e.i.e\; \; \; \frac{1}{4\pi \varepsilon _{0}}\frac{\left ( Ze \right )e}{r^{2}}=\frac{mv^{2}}{r} ....(i)\\ Also\; \; \; mvr=\frac{nh}{2\pi} .........(ii)

From equation (i) and (ii) radius of nth orbit

r_{n}=\frac{n^{2}h^{2}}{4 \pi^{2}kZme^{2}}=\frac{n^{2}h^{2}\varepsilon _{0}}{\pi m Ze^{2}}=0.53\frac{n^{2}}{Z}A^{\circ}\; \; \; \; \; \; \left ( k=\frac{1}{4\pi \varepsilon _{0}} \right )\\ \\ \Rightarrow r_{n}\alpha \frac{n^{2}}{Z}\; or\; r_{n}\alpha \frac{1}{Z}\\ r_{n}=a_{0}\frac{n^{2}}{Z}, \; where\; a_{0}=the \; Bohr\; radius=53pm

The atomic number (Z) of lithium is 3

As r_{a}=a_{0}\frac{n^{2}}{Z}

Therefore the radius of Li++ ion in-ground state on the basis of Bohr's model, will be about \frac{1}{3} times to that of Bohr radius.

Therefore the radius of lithium ion is near r= \frac{53}{3}\approx 18 pm

 

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infoexpert24

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