Question

The ratio of the difference in energy between the first and second Bohr orbit to that between the second and the third Bohr’s orbit is:
A. $\dfrac{5}{{27}}$
B. $\dfrac{1}{3}$
C. $\dfrac{4}{9}$
D. $\dfrac{{27}}{5}$

Answer 1

In order to solve this question we need to understand electron distribution in atoms. So electron configuration around the atom nucleus was first proposed by Niels Bohr, according to him the electron around the atom nucleus is distributed in orbitals around atoms. Also the angular momentum of that orbital is integral multiple of $\dfrac{h}{{2\pi }}$ . So in this question we would first see the relation between the energy of the orbital and principal quantum number “n”. Then we would calculate each required energy level and then we would calculate the difference.

Complete step by step answer:
Let the orbital’s principal quantum number be, “n”. So we know angular momentum is given by, $L = n\dfrac{h}{{2\pi }}$. Also we know velocity of electron in ${n^{th}}$ orbital is given by,
$v = \dfrac{Z}{n}(\dfrac{{{e^2}}}{{2h{\varepsilon _0}}})$
So kinetic energy is given by, $K = \dfrac{1}{2}m{v^2}$
Putting value of velocity is,
$K = \dfrac{1}{2}m{\\{ \dfrac{Z}{n}(\dfrac{{{e^2}}}{{2h{\varepsilon _0}}})\\} ^2}$
$\Rightarrow K = \dfrac{{{Z^2}}}{{{n^2}}}(\dfrac{{m{e^4}}}{{8{h^2}{\varepsilon _0}^2}})$
Since we know total energy is given by, $E = - K$
Putting value we get,
$E = - \dfrac{{{Z^2}}}{{{n^2}}}(\dfrac{{m{e^4}}}{{8{h^2}{\varepsilon _0}^2}})$

Putting value of all constant we get,
$E = - \dfrac{{{Z^2}}}{{{n^2}}}\\{ \dfrac{{(9.1 \times {{10}^{ - 31}}){{(1.6 \times {{10}^{ - 19}})}^4}}}{{8{{(6.626 \times {{10}^{ - 34}})}^2}{{(8.854 \times {{10}^{ - 12}})}^2}}}\\}$
$\Rightarrow E = - 13.6\dfrac{{{Z^2}}}{{{n^2}}}eV$
So energy of $n = 1$ orbit is given by,
${E_1} = - 13.6\dfrac{{{Z^2}}}{{{{(1)}^2}}}eV$
$\Rightarrow {E_1} = - 13.6{Z^2}eV$
So energy of $n = 2$ orbit is given by,
${E_2} = - 13.6\dfrac{{{Z^2}}}{{{{(2)}^2}}}eV$
$\Rightarrow {E_2} = - 3.4{Z^2}eV$
So energy of $n = 3$ orbit is given by,
${E_3} = - 13.6\dfrac{{{Z^2}}}{{{{(3)}^2}}}eV$
$\Rightarrow {E_3} = - 1.51{Z^2}eV$

So difference between first and second is given as,
$\Delta E = {E_2} - {E_1}$
$\Rightarrow \Delta E = ( - 3.4{Z^2}eV) - ( - 13.6{Z^2}eV)$
$\Rightarrow \Delta E = ( - 3.4{Z^2} + 13.6{Z^2})eV$
$\Rightarrow \Delta E = 10.2{Z^2}eV$
So difference between first and second is given as,
$\Delta E' = {E_3} - {E_2}$
$\Rightarrow \Delta E' = ( - 1.51{Z^2}eV) - ( - 3.4{Z^2}eV)$
$\Rightarrow \Delta E' = ( - 1.51{Z^2} + 3.4{Z^2})eV$
$\Rightarrow \Delta E' = 1.89{Z^2}eV$
So the ratio of the difference in energy between the first and second Bohr orbit to that between the second and the third Bohr’s orbit is,
$\dfrac{{\Delta E}}{{\Delta E'}} = \dfrac{{10.2}}{{1.89}}$
$\Rightarrow \dfrac{{\Delta E}}{{\Delta E'}} = 5.39$
$\therefore \dfrac{{\Delta E}}{{\Delta E'}} = \dfrac{{27}}{5}$

So the correct option is D.

Note: It should be remembered that although Bohr atomic configuration was satisfactory but it faces various problems, finally after quantum mechanics calculation, we get to know that when electron is treated when wave function, it produces several other results such as electron distributed in orbitals instead of shell and each orbital is characterized by four quantum numbers.