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The difference between the radii of 3rd and 4th orbits of Li^{2+} is \Delta R_{1}. The difference between the radii of 3rd and 4th orbits of He^{+} is \Delta R_{2}. Ratio \Delta R_{1}:\Delta R_{2}  is  :
Option: 1 8 : 3
Option: 2 3 : 8
Option: 3 2 : 3
Option: 4 3 : 2

Answers (1)

best_answer

We know the formula,

The radius of nth Bohr's Orbit

\mathrm{r = 0.529\times\frac{n^2}{Z}}\mathrm{\ A^{\circ}}

Where
n = principal quantum number of orbit.
Z = Atomic number

Now, 

The difference between the radii of 3rd and 4th orbits of Li^{2+} is \Delta R_{1}. 

\mathrm{\left(R_{4}-R_{3}\right)_{L i^{+2}}=\frac{0.529}{3}\left\{4^{2}-3^{2}\right\}=\Delta R_{1}}

 

The difference between the radii of 3rd and 4th orbits of He^{+} is \Delta R_{2}. 

\left(\mathrm{R}_{4}-\mathrm{R}_{3}\right)_{\mathrm{He}^{+2}}=\frac{0.529}{2}\left\{4^{2}-3^{2}\right\}=\Delta \mathrm{R}_{2}

Ratio \Delta R_{1}:\Delta R_{2}  is  

\frac{\Delta \mathrm{R}_{1}}{\Delta \mathrm{R}_{2}}=\frac{\left(\mathrm{R}_{4}-\mathrm{R}_{3}\right)_{\textup{Li}^{2+}}}{\left(\mathrm{R}_{4}-\mathrm{R}_{3}\right)_{\mathrm{He}^{+}}}=\frac{\frac{4^{2}}{3}-\frac{3^{2}}{3}}{\frac{4^{2}}{2}-\frac{3^{2}}{2}}=\frac{7 / 3}{7 / 2}=\frac{2}{3}

Therefore, the correct option is (3).

Posted by

Kuldeep Maurya

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