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Question: Given \[{r_{n + 1}} - {r_{n - 1}} = 2{r_n}\] ​, where \[{r_n}\] ​, \[{r_{n - 1}}\] ​, \[{r_{n + 1}}\...

Given rn+1−rn−1=2rn{r_{n + 1}} - {r_{n - 1}} = 2{r_n} ​, where rn{r_n} ​, rn−1{r_{n - 1}} ​, rn+1{r_{n + 1}} ​ are Bohr radius for hydrogen atom in nth, (n−1)th, (n+1)th shells respectively. The value of n is:

Explanation

Solution

In the question, we have an algebraic equation involving the Bohr Radius. The Bohr radius is the most probable distance between the nucleus of the hydrogen atom and the electron revolving around it in its ground state. The formula used for calculating the Bohr radius for hydrogen and another hydrogen-like species are:
The radius of the orbit: rn=0.529n2ZA∘  {r_n} = 0.529\dfrac{{{n^2}}}{Z}{A^ \circ }\;
Where n is the principal quantum number of the orbit.
And Z is the atomic number.

Complete step by step answer:
Here, the given equation is: rn+1−rn−1=2rn{r_{n + 1}} - {r_{n - 1}} = 2{r_n}
Now, by using the formula for finding the Bohr radius of orbital i.e.
The radius of the orbit: rn=0.529n2ZA∘  {r_n} = 0.529\dfrac{{{n^2}}}{Z}{A^ \circ }\;
We put the value of rn{r_n} in the above equation:
0.529(n+1)2Z−0.529(n−1)2Z  =2×0.529×n2Z    0.529\dfrac{{{{(n + 1)}^2}}}{Z} - 0.529\dfrac{{{{(n - 1)}^2}}}{Z}\; = 2 \times 0.529 \times \dfrac{{{n^2}}}{Z}\;\;
Here, the value of Z is 1 because we are calculating for the hydrogen atom as mentioned in the question statement.
Thus, the equation becomes:
(n+1)2−(n−1)2=2(n)2{\left( {n + 1} \right)^2} - {\left( {n - 1} \right)^2} = 2{\left( n \right)^2}
Using the mathematical equation(a+b)2=a2+2ab+b2{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}, we get:
(n2+2n+1)−(n2−2n+1)=2(n)2\left( {{n^2} + 2n + 1} \right) - \left( {{n^2} - 2n + 1} \right) = 2{\left( n \right)^2}
4n=2n24n = 2{n^2}
n=2n = 2

hence, the value of n i.e. the principal quantum number is 2 .

Additional information:
Bohr’s model of the hydrogen atom is based on three postulates:
An electron revolves around the nucleus in a specified circular orbit,
An electron’s angular momentum in the orbit is quantized,
A photon is emitted whenever the electron in the excited state comes back to its ground state.
In postulate 1 Bohr considered the classical approach and the second and third postulate is based on the quantization principle, this makes Bohr’s model a semiclassical model.

Note: Originally, the formula for Bohr radius is rn=aBn2Z Ao  {r_n} = {a_B}\dfrac{{{n^2}}}{Z}{\text{ }}{A^{o\;}}
Where aB{a_B} is the constant, whose expression is:
aB=h24π2mekqe2=0.529×10−10maB = \dfrac{{{h^2}}}{{4{\pi ^2}{m_e}kq_e^2}} = 0.529 \times {10^{ - 10}}m
Where h is the planck's constant =  6.626×10−34  m2kg/s = \;6.626 \times {10^{ - 34\;}}{m^2}kg/s
me{m_e} is the rest mass of electron =9.109×10−31kilograms = 9.109 \times {10^{ - 31}}kilograms
kk is the coulomb’s constant =8.988×109Nm2C−2 = 8.988 \times {10^9}N{m^2}{C^{ - 2}}
qe{q_e} is the electron charge =1.602×10−19coulombs = 1.602 \times {10^{ - 19}}coulombs