Question

A doubly ionized lithium atom is hydrogen-like with atomic number 3.
(a) Find the wavelength of the radiation required to excite the electron in \[Li^{++}\] from the first to the third Bohr orbit. (Ionization energy of the hydrogen atom equals \[13.6\,{\text{eV}}\])
(b) How many spectral lines are observed in the emission spectrum of the above excited system?
A. \[114.7\,\mathop {\text{A}}\limits^{\text{o}} \], three lines
B. \[113.7\,\mathop {\text{A}}\limits^{\text{o}} \], three lines
C. \[115.7\,\mathop {\text{A}}\limits^{\text{o}} \], two lines
D. \[116.7\,\mathop {\text{A}}\limits^{\text{o}} \], one line

Answer 1

Hint:Use the formula for determining the energy of the electron in nth orbit. Determine the energies of first and third Bohr orbit and calculate the energy required to excite the electron. Use the formula for energy of the photon to determine the wavelength of the photon. Check the number of ways by which the excited electron can come to its ground state which is the first Bohr orbit.

Formulae used:
The energy \[{E_n}\] of an electron of hydrogen atom in nth orbit is given by
\[{E_n} = - 13.6\dfrac{{{z^2}}}{{{n^2}}}\,{\text{eV}}\] …… (1)
Here, \[z\] is the atomic number of the atom and \[n\] is the principal quantum number.
The energy \[E\] of a photon is given by
\[E = \dfrac{{hc}}{\lambda }\] …… (2)
Here, \[h\] is the Planck’s constant, \[c\] is the speed of light and \[\lambda \] is the wavelength of a photon.

Complete step by step answer:
We have given that a doubly ionized lithium atom with atomic number 3 behaves like a hydrogen atom.
\[z = 1\]
(a) When the lithium atom is double ionized, only one electron remain in the orbit of the lithium atom.We need to determine the energy required for the electron in the first orbit to excite to the third orbit.We can determine the energy of the electron in the first and third Bohr orbit using equation (1).
Substitute \[1\] for \[n\] and \[3\] for \[z\] in equation (1).
\[{E_1} = - 13.6\dfrac{{{{\left( 3 \right)}^2}}}{{{{\left( 1 \right)}^2}}}\,{\text{eV}}\]
\[ \Rightarrow {E_1} = - 122.4\,{\text{eV}}\]
Substitute \[3\] for \[n\] and \[3\] for \[z\] in equation (1).
\[{E_3} = - 13.6\dfrac{{{{\left( 3 \right)}^2}}}{{{{\left( 3 \right)}^2}}}\,{\text{eV}}\]
\[ \Rightarrow {E_3} = - 13.6\,{\text{eV}}\]
The energy \[\Delta E\] required to excite the electron from the first orbit to the third orbit is equal to the difference between the energy \[{E_3}\] of the third orbit and energy \[{E_1}\] of the first orbit.
\[\Delta E = {E_3} - {E_1}\]
Substitute \[ - 13.6\,{\text{eV}}\] for \[{E_3}\] and \[ - 122.4\,{\text{eV}}\] for \[{E_1}\] in the above equation.
\[\Delta E = \left( { - 13.6\,{\text{eV}}} \right) - \left( { - 122.4\,{\text{eV}}} \right)\]
\[ \Rightarrow \Delta E = 108.8\,{\text{eV}}\]
\[ \Rightarrow \Delta E = \left( {108.8\,{\text{eV}}} \right)\left( {\dfrac{{1.6 \times {{10}^{ - 19}}\,{\text{J}}}}{{1\,{\text{eV}}}}} \right)\]
\[ \Rightarrow \Delta E = 174.08 \times {10^{ - 19}}\,{\text{J}}\]
Let us now determine the wavelength of the photon required to excite the electron.
Rewrite equation (2) for the energy required to excite electrons.
\[\Delta E = \dfrac{{hc}}{\lambda }\]
\[ \Rightarrow \lambda = \dfrac{{hc}}{{\Delta E}}\]
Substitute \[6.6 \times {10^{ - 34}}\,{\text{J}} \cdot {\text{s}}\] for \[h\], \[3 \times {10^8}\,{\text{m/s}}\] for \[c\] and \[174.08 \times {10^{ - 19}}\,{\text{J}}\] for \[\Delta E\] in the above equation.
\[ \Rightarrow \lambda = \dfrac{{\left( {6.6 \times {{10}^{ - 34}}\,{\text{J}} \cdot {\text{s}}} \right)\left( {3 \times {{10}^8}\,{\text{m/s}}} \right)}}{{174.08 \times {{10}^{ - 19}}\,{\text{J}}}}\]
\[ \Rightarrow \lambda = 113.7 \times {10^{ - 10}}\,{\text{m}}\]
\[ \therefore \lambda = 113.7\,\mathop {\text{A}}\limits^{\text{o}} \]

Hence, the wavelength required to excite the electron is \[113.7\,\mathop {\text{A}}\limits^{\text{o}} \].

(b) Now, we have to determine the number of spectral lines observed in the emission spectrum of the excited electron in the third Bohr orbit while coming down to the first Bohr orbit.The number of possible ways by which the excited electron can come to first orbit are as follows:
1.From third orbit to first orbit (\[3 \to 1\])
2.From third orbit to second orbit and then to first orbit (\[3 \to 2\], \[2 \to 1\])

Therefore, the number of the spectral lines observed in the emission spectra are three.Hence, the correct option is B.

Note:One should not forget to convert the energy difference obtained from electronvolt to joule as the further calculations are in the SI system of units. Also, while determining the number of spectral lines in the emission spectra, one should keep in mind that the ground state of the electron is the first Bohr orbit and not the orbit with principle quantum number 0.