Question

What total amount of energy in calories would be required to shift all the electrons from the first Bohr orbit to the sixth Bohr orbit in 1 mole of hydrogen? Through what distance would each electron have to move? What frequency of radiation would be emitted if the electrons returned to their initial state?

Answer 1

Hint: In the modern atomic structure, we use Bohr orbit, which is also termed as energy states. Two Bohr orbits are distinct in terms of their radii and energies. The electrons residing in the same Bohr orbit have the same amount of energy. Thus, the energy-difference of two electrons residing in two different orbits is the same as the energy difference between two corresponding orbits.

Complete answer:
In the year of $1913$, Neils Bohr proposed a hypothesis where he first introduced the idea of Bohr orbit. According to the hypothesis, an electron can only reside in a particular energy state or Bohr orbit. From this hypothesis, we have-${E_n} = \dfrac{{ - 13.6}}{{{n^2}}}eV$
Where,
${E_n}$ is the energy of ${n^{th}}$-Bohr orbit
$- 13.6eV$ is the energy of the first Bohr orbit
And,
${r_n} = {n^2} \times 0.53{A^ \circ }$
Where,
${r_n}$ is the radius of ${n^{th}}$-Bohr orbit
$0.53{A^ \circ }$ is the radius of the first Bohr orbit
Therefore, the energy required to shift an electron from ${1^{st}}$ orbit to ${6^{th}}$ orbit is-
$\Delta E = {E_6} - {E_1}$
$\Delta E = - 13.6 \times \left( {\dfrac{1}{{{6^2}}} - \dfrac{1}{{{1^2}}}} \right)$
$\Delta E = - 13.6 \times \left( {\dfrac{1}{{36}} - 1} \right)$
$\Delta E = - 13.6 \times \left( {\dfrac{{ - 35}}{{36}}} \right)$
$\Delta E = 13.22eV$
We convert it into calories-
$\Delta E = 13.22 \times 1.6 \times {10^{ - 19}}J$
$\Delta E = 2.1152 \times {10^{ - 18}}J$
Now we convert it into calorie-
$\Delta E = \dfrac{{2.1152 \times {{10}^{ - 18}}}}{{4.2}}cal$
$\Delta E = 5.036 \times {10^{ - 19}}cal$
For $1mol$Hydrogen, the total energy is-$\Delta {E_{tot}} = 5.036 \times {10^{ - 19}} \times 6.022 \times {10^{23}}cal$
$= 303.28 \times {10^3}cal$
The distance each electron will have to move due to the shift is-
$\Delta r = 0.53 \times \left( {{6^2} - {1^2}} \right){A^ \circ }$
$= \left( {0.53 \times 35} \right){A^ \circ }$
$= 18.55{A^ \circ }$
We know the frequency of the emitted radiation can be given as-
$\Rightarrow \nu = \dfrac{{\Delta E}}{h}$
Where,
$h$ is the Planck's constant
$\nu$ is the frequency of the radiation
We put $\Delta E = 2.1152 \times {10^{ - 18}}J$ and $h = 6.625 \times {10^{ - 34}}J.s$-
$\therefore \nu = \dfrac{{2.1152 \times {{10}^{ - 18}}}}{{6.625 \times {{10}^{ - 34}}}}Hz$
$= 3.19 \times {10^{15}}Hz$

Note: Bohr’s hypothesis can only be applicable to one-electron systems such as Hydrogen atom or Hydrogen-like atoms such as $H{e^{2 + }}$, $L{i^{3 + }}$. Bohr’s theory also has some drawbacks, like it can not explain emission spectra of multi-electrons systems. It also fails to cope up with de Broglie’s hypothesis of the dual nature of matter.