The difference between the radii of 3^{rd} an | vedclass

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(D) The radius of an orbit is given by $r_{n} = a_{0} 	imes \frac{n^{2}}{Z}$,where $a_{0}$ is the Bohr radius,$n$ is the orbit number,and $Z$ is the

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$2:3$

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(D) The radius of an orbit is given by $r_{n} = a_{0} 	imes \frac{n^{2}}{Z}$,where $a_{0}$ is the Bohr radius,$n$ is the orbit number,and $Z$ is the atomic number.For $Li^{2+}$,$Z = 3$. Thus,$\Delta R_{1} = r_{4} - r_{3} = a_{0} 	imes \frac{4^{2} - 3^{2}}{3} = a_{0} 	imes \frac{16 - 9}{3} = \frac{7}{3} a_{0}$.For $He^{+}$,$Z = 2$. Thus,$\Delta R_{2} = r_{4} - r_{3} = a_{0} 	imes \frac{4^{2} - 3^{2}}{2} = a_{0} 	imes \frac{16 - 9}{2} = \frac{7}{2} a_{0}$.The ratio $\Delta R_{1} : \Delta R_{2} = \frac{7/3}{7/2} = \frac{2}{3}$.

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

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