Question

Find the energy released when 2.0 mole of atoms of Hydrogen undergo transition, giving a spectral line of lowest energy in the visible region of its atomic spectra (in KJ).

Answer 1

Hint: Formula for defining the energy levels of a Hydrogen atom are given by the equation
\[E{\text{ }} = {\text{ }}\dfrac{{ - {E_0}}}{{{n^2}}},\] where \[{E_0}\; = {\text{ }}13.6{\text{ }}eV\;{\text{ and 1eV = 1}}{\text{.602}} \times {\text{1}}{{\text{0}}^{ - 19}}Joules\] and n = 1,2,3… and so on.
The energy is expressed as a negative number because it takes that much energy to unbind or ionize the electron from the nucleus.
Rydberg Formula is for the subtraction of two energy levels.
${E_n} = {E_o}\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right) \\
  where,{\text{ }}{{\text{n}}_{\text{1}}}{\text{ < }}{{\text{n}}_2} \\
$

Complete step by step answer:
Formula for the electron energy levels in hydrogen:
$\Rightarrow$ \[{E_n}{\text{ }} = - {R_H}{Z^2}\left( {\dfrac{1}{{{n^2}}}{\text{ }}} \right)\]for n = 1, 2, 3, 4, . . . .
 Where,
$\Rightarrow$ \[{R_H}{\text{ }} = {\text{ }}2.179{\text{ }}x{\text{ }}{10^{ - 18}}Joules\] = Rydberg constant
Z = atomic no. of the atom
n = Principal Quantum Number
The energy change associated with a transition between electron energy levels can be given as
$ \vartriangle E = {E_{final}} - {E_{initial}} \\
  \vartriangle E = {R_H}{Z^2}\left[ {\dfrac{1}{{n_i^2}} - \dfrac{1}{{n_f^2}}} \right] \\
$
And in the Balmer series the photons emitted by transition are in the visible region.
For balmer the electron transitioning from \[n\; \geqslant 3{\text{ }}to{\text{ }}\;n\; = {\text{ }}2.\]
And for the lowest energy \[n\; = 3{\text{ and }}\;n\; = {\text{ }}2.\]
$ \vartriangle E = {E_3} - {E_2} \\
  \vartriangle E = {R_H}{Z^2}\left[ {\dfrac{1}{{n_2^2}} - \dfrac{1}{{n_3^2}}} \right] \\
\Rightarrow {R_H}{Z^2}\left[ {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}}} \right] \\
\Rightarrow {R_H}{Z^2}\left[ {\dfrac{1}{4} - \dfrac{1}{9}} \right] \\
  \vartriangle E = 1312 \times \dfrac{5}{{36}} = 182.2{\text{ KJ mo}}{{\text{l}}^{ - 1}} \\
$
For 2 mole,
$\Rightarrow$ $\vartriangle E = 182.2 \times 2 = 364.4KJ$

Note:
The masses of photons are inversely proportional to the wavelength but directly proportional to frequency of the photon. Brackett and Balmer series occur in the infra-red and visible region of the spectrum respectively.