Answer
405k+ views
Hint: Apply bohr’s postulates to find the relationship between different physical quantities of atomic orbits, also you may need to apply law of energy conservation to find some physical quantities like velocity of electrons.
Complete answer:
Let mass of electron be m
It moves with speed v and radius of ${n^{th}}$orbit is ${r_n}$
Now according to bohr’s postulate (angular momentum is integral multiple of $\hbar $)
We have
$mvr = n\hbar $
As centripetal force is due to electrostatic force,therefore we can say;
$centripetal\,force = \dfrac{{m{v^2}}}{{{r_n}}} = k\dfrac{{Ze.e}}{{{r_n}^2}}$
Solving both these equations we have
Radius ${r_n} = \dfrac{{2{n^2}{\hbar ^2}}}{{mZ{e^2}}}$
${r_n} \propto \dfrac{{{n^2}}}{Z}$
Relative change in radius of two consecutive radius will be
\[
\dfrac{{{r_{n + 1}} - {r_n}}}{{{r_n}}} = \dfrac{{\dfrac{{{{(n + 1)}^2} - {n^2}}}{Z}}}{{\dfrac{{{n^2}}}{Z}}} \\
\Rightarrow\dfrac{{{r_{n + 1}} - {r_n}}}{{{r_n}}}= \dfrac{{\dfrac{{2n + 1}}{Z}}}{{\dfrac{{{n^2}}}{Z}}}\]
For n>>1 we can say
Relative change in radius of of consecutive orbital is,
\[\dfrac{{{r_{n + 1}} - {r_n}}}{{{r_n}}} \propto \dfrac{1}{n}\]
Also it does not depend upon Z
So option-A and option-B are correct
We know that
Energy $E \propto \dfrac{1}{{{n^2}}}$
So relative change in energy between two consecutive orbitals will be
$\dfrac{{{E_n} - {E_{n + 1}}}}{{{E_n}}} = \dfrac{{\dfrac{1}{{{n^2}}} - \dfrac{1}{{{{\left( {n + 1} \right)}^2}}}}}{{\dfrac{1}{{{n^2}}}}}\\
\therefore\dfrac{{{E_n} - {E_{n + 1}}}}{{{E_n}}} = \dfrac{{2n + 1}}{{{n^2}}}$
For larger value of n we can say it is proportional to $\dfrac{1}{n}$
So option-C is incorrect. Similarly we can show that relative change in angular momenta of two consecutive orbitals is proportional to $\dfrac{1}{n}$.
Hence Options A,B and D are correct.
Note: Here its asking relative change in each and every it’s not asking just change so you will have to divide with initial value to get relative change else, also you will have to use approximations as value of n is much much larger.
Complete answer:
Let mass of electron be m
It moves with speed v and radius of ${n^{th}}$orbit is ${r_n}$
Now according to bohr’s postulate (angular momentum is integral multiple of $\hbar $)
We have
$mvr = n\hbar $
As centripetal force is due to electrostatic force,therefore we can say;
$centripetal\,force = \dfrac{{m{v^2}}}{{{r_n}}} = k\dfrac{{Ze.e}}{{{r_n}^2}}$
Solving both these equations we have
Radius ${r_n} = \dfrac{{2{n^2}{\hbar ^2}}}{{mZ{e^2}}}$
${r_n} \propto \dfrac{{{n^2}}}{Z}$
Relative change in radius of two consecutive radius will be
\[
\dfrac{{{r_{n + 1}} - {r_n}}}{{{r_n}}} = \dfrac{{\dfrac{{{{(n + 1)}^2} - {n^2}}}{Z}}}{{\dfrac{{{n^2}}}{Z}}} \\
\Rightarrow\dfrac{{{r_{n + 1}} - {r_n}}}{{{r_n}}}= \dfrac{{\dfrac{{2n + 1}}{Z}}}{{\dfrac{{{n^2}}}{Z}}}\]
For n>>1 we can say
Relative change in radius of of consecutive orbital is,
\[\dfrac{{{r_{n + 1}} - {r_n}}}{{{r_n}}} \propto \dfrac{1}{n}\]
Also it does not depend upon Z
So option-A and option-B are correct
We know that
Energy $E \propto \dfrac{1}{{{n^2}}}$
So relative change in energy between two consecutive orbitals will be
$\dfrac{{{E_n} - {E_{n + 1}}}}{{{E_n}}} = \dfrac{{\dfrac{1}{{{n^2}}} - \dfrac{1}{{{{\left( {n + 1} \right)}^2}}}}}{{\dfrac{1}{{{n^2}}}}}\\
\therefore\dfrac{{{E_n} - {E_{n + 1}}}}{{{E_n}}} = \dfrac{{2n + 1}}{{{n^2}}}$
For larger value of n we can say it is proportional to $\dfrac{1}{n}$
So option-C is incorrect. Similarly we can show that relative change in angular momenta of two consecutive orbitals is proportional to $\dfrac{1}{n}$.
Hence Options A,B and D are correct.
Note: Here its asking relative change in each and every it’s not asking just change so you will have to divide with initial value to get relative change else, also you will have to use approximations as value of n is much much larger.
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