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Taking the Bohr radius as \mathrm{a_0= 53\: pm, } the radius of \mathrm{Li^{++} } ion in its ground state, on the basis of Bohr’s model, will be about
 

Option: 1

\mathrm{53 \: pm}
 


Option: 2

\mathrm{27\: pm}


Option: 3

\mathrm{18\: pm}


Option: 4

\mathrm{13\: pm}


Answers (1)

best_answer

On the basis of Bohr's model, \mathrm{r=\frac{n^2 h^2}{4 \pi^2 m K Z e^2}=a_0 \frac{n^2}{Z} \: }

\mathrm{For \: \: \mathrm{Li}^{++}ion, Z=3, n=1 } for ground state.

Given,\mathrm{ a_0=53 \mathrm{pm}}

\mathrm{ \therefore \quad r=\frac{53 \times 1^2}{3}=18 \mathrm{pm} }

Hence option 3 is correct.
 

Posted by

avinash.dongre

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