Question

Calculate ratio of Bohr's radius of second orbit of li^2+ ions to the bohr's radius of third orbit of He^+ ion

Answer 1

Hi , there !!!


here is your answer !!!

Bohr's radius of second orbit is given by


r = 0.529 × n² / z

0.529 × 2² / 3

=> 0.529 × 4 / 3





now ,

Bohr's third radius of He2+


is .529× n² / z


0529 × 9 / 2


hence the ratio come out to be


4 / 3 / 9/2

=> 8 / 27


hope it helps you !!!

thanks !!!

Ranjan kumar

Answer 2

$\huge\underbrace\pink {\dag Final \: Answer:- \dag}$  

$8:27$

$\huge\underbrace\red{\dag To \: Find:-\dag}$

ratio of Bohr's radius of second orbit of $Li^{2+}$ ions to the bohr's radius of third orbit of $He^{1+}$ ion

$\huge\underbrace\blue{\dag Formula Used:-\dag}$

$R_n =\frac{n^{2}}{z} \times 0.529 \AA[\tex]</p><p></p><p>[tex]\huge\underbrace\green{\dag Things \: to \:know:-\dag}$

$R_n =$ radius

n= number of orbit  

z= atomic number  

$\huge\underbrace\blue{\dag ATQ:-\dag}$

$n_1=2 \:\:\:\: z_1=3 \\n_2=3 \:\:\: z_2=2 \\\AA = 10^{-10}$

$\huge\underbrace\pink{\dag Solution:- \dag}$

Bohr's radius of second orbit of $Li^{2+}$ ions ⇒ $R_L_i$

$R_L_i=\frac{n_1^{2}}{z_1} \times 0.529 \AA \\ R_L_i=\frac{2^{2}}{3} \times 0.529 \times 10^{-10} \\ R_L_i=\frac{4}{3} \times 0.529 \times 10^{-10} \\Or  \\R_L_i= 0.7053 \times 10^{-10}$

Bohr's radius of third orbit of $He^{1+}$ ion ⇒ $R_H_e$

$R_H_e =\frac{n_2^{2}}{z_2} \times 0.529 \AA \\  R_H_e=\frac{3^{2}}{2} \times 0.529  \times 10^{-10}  \\  R_H_e=\frac{9}{2} \times 0.529  \times 10^{-10}  \\Or\\  R_H_e=2.3805 \times 10^{-10}$

so ratio will be

$=\frac{ R_L_i }{ R_H_e } \\=\frac{\frac{4}{3} \times 0.529 \times 10^{-10} }{\frac{9}{2} \times 0.529  \times 10^{-10}  } \\=\frac{\frac{4}{3}} {\frac{9}{2}} \\ ={\frac{8}{27} }$

hence their ratio will be $\boxed{8:27}$