In a collection of \[{{H}}\] atom, electrons make transition from \[{{{5}}^{{{th}}}}\] excited state to \[{{{5}}^{{{th}}}}\] excited state, then the maximum number of different types of photons observed are;
A. \[{{3}}\]
B. \[{{4}}\]
C. \[{{6}}\]
D. \[{{15}}\]
Answer
626.1k+ views
Hint: \[{{{n}}^{{{th}}}}\]excited state is equal to the \[{{(n + 1)}}\] transition line, first excited state can be considered as \[{{{n}}_{{2}}}\] and second excited state can be considered as \[{{{n}}_{{1}}}\].Substituted these values in number of photons lin formula we get maximum observed photons in the spectrum.
Complete step by step answer:
Maximum number of lines can be calculated by the following formula.
\[{\text{Number of lines = }}\dfrac{{{{(}}{{{n}}_{{2}}}{{ - }}{{{n}}_{{1}}}{{)(}}{{{n}}_{{2}}}{{ - }}{{{n}}_{{1}}}{{ + 1)}}}}{{{2}}}\]……………………………..(1)
From the given,
\[{{{5}}^{{{th}}}}\] excited state = \[{6^{{{th}}}}\]transition line = \[{{{n}}_{{2}}}\]
\[{{{2}}^{{{nd}}}}\]excited state = \[{{{3}}^{{{rd}}}}\] transition line = \[{{{n}}_{{1}}}\]
Substitute these values in equation (1)
\[
\Rightarrow \dfrac{{(6 - 3)(6 - 3 + 1)}}{2} \\
\Rightarrow \dfrac{{(3)(4)}}{2} \\
\Rightarrow (3) \times (2) \\
\Rightarrow 6 \\
\]
So, the maximum number of lines of different photons will be observed six.
The number of emissions lines are as follows.
\[6 \to 3,6 \to 4,6 \to 5,5 \to 3,5 \to 4,4 \to 3\]
Hence, the correct option is (C).
Additional Information:
For this type of question, one should remember that they are asking the type of photons observed. If they ask about spectral lines when electron jump from \[{{{n}}^{{{th}}}}\] level to the ground state then we have to use the formula of spectral lines from transition to ground state.
\[\dfrac{{{{{N}}_{{2}}}{{(}}{{{N}}_{{2}}}{{ - 1)}}}}{{{2}}}\]
In hydrogen atom absorption spectra is obtained by giving energy to \[{{H}}\]- atom and passing the released light energy through prism for obtaining spectra.
Note:
When an electron jumps from higher energy orbit to lower energy orbit, it emits radiation with various spectral lines. In a \[{{H}}\]-atom when an electron makes a transition from an excited state to the ground state its kinetic energy increases while potential and total energy decreases.
Complete step by step answer:
Maximum number of lines can be calculated by the following formula.
\[{\text{Number of lines = }}\dfrac{{{{(}}{{{n}}_{{2}}}{{ - }}{{{n}}_{{1}}}{{)(}}{{{n}}_{{2}}}{{ - }}{{{n}}_{{1}}}{{ + 1)}}}}{{{2}}}\]……………………………..(1)
From the given,
\[{{{5}}^{{{th}}}}\] excited state = \[{6^{{{th}}}}\]transition line = \[{{{n}}_{{2}}}\]
\[{{{2}}^{{{nd}}}}\]excited state = \[{{{3}}^{{{rd}}}}\] transition line = \[{{{n}}_{{1}}}\]
Substitute these values in equation (1)
\[
\Rightarrow \dfrac{{(6 - 3)(6 - 3 + 1)}}{2} \\
\Rightarrow \dfrac{{(3)(4)}}{2} \\
\Rightarrow (3) \times (2) \\
\Rightarrow 6 \\
\]
So, the maximum number of lines of different photons will be observed six.
The number of emissions lines are as follows.
\[6 \to 3,6 \to 4,6 \to 5,5 \to 3,5 \to 4,4 \to 3\]
Hence, the correct option is (C).
Additional Information:
For this type of question, one should remember that they are asking the type of photons observed. If they ask about spectral lines when electron jump from \[{{{n}}^{{{th}}}}\] level to the ground state then we have to use the formula of spectral lines from transition to ground state.
\[\dfrac{{{{{N}}_{{2}}}{{(}}{{{N}}_{{2}}}{{ - 1)}}}}{{{2}}}\]
In hydrogen atom absorption spectra is obtained by giving energy to \[{{H}}\]- atom and passing the released light energy through prism for obtaining spectra.
Note:
When an electron jumps from higher energy orbit to lower energy orbit, it emits radiation with various spectral lines. In a \[{{H}}\]-atom when an electron makes a transition from an excited state to the ground state its kinetic energy increases while potential and total energy decreases.
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