
All the electrons in a sample of the hydrogen atom are excited to \[n{\text{ }} = {\text{ }}7\] level. The number of emission spectral lines possible are:
A. 12
B. 15
C. 21
D. 18
Answer
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Hint:To calculate the number of emission spectral lines possible.
\[N = \dfrac{{\left( {{n_2} - {n_1}} \right)\left( {{n_2} - {n_1} + 1} \right)}}{2}\]… …(1)
Here, \[{n_1}\] and \[{n_2}\] represent the ground state energy level and excited state energy level. N represents the number of emission spectral lines possible.
Complete answer:
The atomic number of hydrogen is 1. Thus, only one electron is present in hydrogen atom.
So, when you say that all the electrons in a sample of the hydrogen atom are excited, you mean that one electron in a sample of the hydrogen atom is excited.
When an electron in a hydrogen atom is excited from \[{n_1}\] to \[{n_2}\] you can use the following expression to calculate the number of emission spectral lines possible.
\[N = \dfrac{{\left( {{n_2} - {n_1}} \right)\left( {{n_2} - {n_1} + 1} \right)}}{2}\]… …(1)
Here, \[{n_1}\] and \[{n_2}\] represent the ground state energy level and excited state energy level. N represents the number of emission spectral lines possible.
One electron in a sample of the hydrogen atom is excited to \[n{\text{ }} = {\text{ }}7\] level.
Here, the ground state energy level \[{n_1} = 1\] and the excited state energy level \[{n_2} = 7\]
Substitute \[{n_1} = 1\] and \[{n_2} = 7\] in the expression (1) and calculate the number of emission spectral lines possible.
\[
N = \dfrac{{\left( {{n_2} - {n_1}} \right)\left( {{n_2} - {n_1} + 1} \right)}}{2} \\
\Rightarrow N = \dfrac{{\left( {7 - 1} \right)\left( {7 - 1 + 1} \right)}}{2}
\]
\[
\Rightarrow N = \dfrac{{6 \times 7}}{2} \\
\Rightarrow N = \dfrac{{42}}{2} \\
\Rightarrow N = 21
\]
The number of emission spectral lines possible are 21.
Hence, the option C. is the correct option.
Note:
In the absorption spectrum, the electron is transferred from lower energy level to higher energy level, as some energy is absorbed during the process. In the emission spectrum, the electron is transferred from higher energy level to lower energy level, as some energy is liberated during the process.
\[N = \dfrac{{\left( {{n_2} - {n_1}} \right)\left( {{n_2} - {n_1} + 1} \right)}}{2}\]… …(1)
Here, \[{n_1}\] and \[{n_2}\] represent the ground state energy level and excited state energy level. N represents the number of emission spectral lines possible.
Complete answer:
The atomic number of hydrogen is 1. Thus, only one electron is present in hydrogen atom.
So, when you say that all the electrons in a sample of the hydrogen atom are excited, you mean that one electron in a sample of the hydrogen atom is excited.
When an electron in a hydrogen atom is excited from \[{n_1}\] to \[{n_2}\] you can use the following expression to calculate the number of emission spectral lines possible.
\[N = \dfrac{{\left( {{n_2} - {n_1}} \right)\left( {{n_2} - {n_1} + 1} \right)}}{2}\]… …(1)
Here, \[{n_1}\] and \[{n_2}\] represent the ground state energy level and excited state energy level. N represents the number of emission spectral lines possible.
One electron in a sample of the hydrogen atom is excited to \[n{\text{ }} = {\text{ }}7\] level.
Here, the ground state energy level \[{n_1} = 1\] and the excited state energy level \[{n_2} = 7\]
Substitute \[{n_1} = 1\] and \[{n_2} = 7\] in the expression (1) and calculate the number of emission spectral lines possible.
\[
N = \dfrac{{\left( {{n_2} - {n_1}} \right)\left( {{n_2} - {n_1} + 1} \right)}}{2} \\
\Rightarrow N = \dfrac{{\left( {7 - 1} \right)\left( {7 - 1 + 1} \right)}}{2}
\]
\[
\Rightarrow N = \dfrac{{6 \times 7}}{2} \\
\Rightarrow N = \dfrac{{42}}{2} \\
\Rightarrow N = 21
\]
The number of emission spectral lines possible are 21.
Hence, the option C. is the correct option.
Note:
In the absorption spectrum, the electron is transferred from lower energy level to higher energy level, as some energy is absorbed during the process. In the emission spectrum, the electron is transferred from higher energy level to lower energy level, as some energy is liberated during the process.
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