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
610.5k+ views
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.
Recently Updated Pages
Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

Find the value of the expression given below sin 30circ class 11 maths CBSE

Difference between physical and chemical change class 11 chemistry CBSE

Two of the body parts which do not appear in MRI are class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

