How many spectral lines are observed in the spectra of ions?
A. 2
B. 4
C. 6
D. 8
Answer
628.8k+ views
Hint:The above problem can be resolved using the concept of the ionizations and the electronic excitation from one state to another state. The electrons possess some definite value of energy at each state known as energy levels. These energy levels, along with the defined energy state, determine the number of emission lines at atomic spectra. Moreover, the mathematical relation for the number of emission spectra is to consider to obtain the desired result.
Complete step by step answer:
The numerical value for the energy levels of the helium are:
-54.4 eV
-13.6 eV
-6.04 eV and so on.
As the helium \[\left( {H{e^ + }} \right)\] ion has its electron in the excited state. Therefore, the spectral lines will occur only when the excitation state of the electron is 2. That is, \[n = 2\].
And for this excitation state, the value of the energy levels is \[10.2\;{\rm{eV}}\]. Moreover there will be ionization of the electron and due to this the electrons receive some amount of energy and get excited to another level that is ,\[n = 4\].
This causes the number of emission lines to be as,
\[{n_1} = \dfrac{{3n}}{2}\]
Substituting the values as,
\[\begin{array}{l}
{n_1} = \dfrac{{3n}}{2}\\
{n_1} = \dfrac{{3 \times 4}}{2}\\
{n_1} = 6
\end{array}\]
Therefore, the number of emission lines is 6 and option (C) is correct.
Note: Try to understand the term "emission lines" and the significant fundamentals regarding the emission, along with the reason for the formation of such lines in the atomic spectra. Moreover, the concept behind this is related to the excitation of electrons and the energy levels of the electrons in each excitation state.
Complete step by step answer:
The numerical value for the energy levels of the helium are:
-54.4 eV
-13.6 eV
-6.04 eV and so on.
As the helium \[\left( {H{e^ + }} \right)\] ion has its electron in the excited state. Therefore, the spectral lines will occur only when the excitation state of the electron is 2. That is, \[n = 2\].
And for this excitation state, the value of the energy levels is \[10.2\;{\rm{eV}}\]. Moreover there will be ionization of the electron and due to this the electrons receive some amount of energy and get excited to another level that is ,\[n = 4\].
This causes the number of emission lines to be as,
\[{n_1} = \dfrac{{3n}}{2}\]
Substituting the values as,
\[\begin{array}{l}
{n_1} = \dfrac{{3n}}{2}\\
{n_1} = \dfrac{{3 \times 4}}{2}\\
{n_1} = 6
\end{array}\]
Therefore, the number of emission lines is 6 and option (C) is correct.
Note: Try to understand the term "emission lines" and the significant fundamentals regarding the emission, along with the reason for the formation of such lines in the atomic spectra. Moreover, the concept behind this is related to the excitation of electrons and the energy levels of the electrons in each excitation state.
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