
Number of possible spectral lines which may be emitted in Brackett series in H-atom if the electrons in ninth excited state returns to ground state are:
[A] 5
[B] 45
[C] 6
[D] 21
Answer
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HINT: For Brackett series the ground state is the fourth energy shell. To solve this question, find out the energy level corresponding to the ninth excited state. Use it to find the possibilities of electrons moving from fourth to that energy level.
COMPLETE STEP BY STEP SOLUTION: We know that spectral emission occurs when an electron transitions from a higher energy state to a lower energy state. The lower energy state is designated as n’ and the higher energy state is designated as n. The spectral lines obtained due to the emissions are grouped into a series according to their lower energy state (n’).
The Brackett series is the emission or absorption lines for H- atoms due to the electron jumps between the fourth energy level and other higher energy levels.
In the question, we are asked the number of possible spectral lines which may be emitted in the Brackett series in H-atom if the electrons in the ninth excited state return to the ground state.
Now for the Brackett series, the lowest energy level i.e. the ground state is n’=4 and the electron is in the ninth excited state.
We know that the first excited state will be the second energy level (as the electrons move from the first energy level to the second).
So, the ninth excited state will be the tenth energy level i.e. n = 10.
So, we have to calculate the number of possible spectral lines when the electron moves from n = 10 to n’ = 4.
So the possibilities are 10 $\to $ 4, 9 $\to $ 4, 8 $\to $ 4, 7 $\to $ 4, 6 $\to $ 4, 5$\to $ 4.
We can see from the above discussion that there are 6 possibilities.
Therefore, the correct answer is option [C] 6.
NOTE: For a hydrogen atom, the spectral lines are given special names.
- For n’=1 the series is called the Lyman series, which lie in the ultraviolet region,
- n’=2 is given the name Balmer series which has a few lines in the visible region,
- n’=3 is the Paschen series lying in the infrared region,
And there are others for n’= 4, 5 and 6 named as Brackett series, Pfund series and Humphrey series respectively.
COMPLETE STEP BY STEP SOLUTION: We know that spectral emission occurs when an electron transitions from a higher energy state to a lower energy state. The lower energy state is designated as n’ and the higher energy state is designated as n. The spectral lines obtained due to the emissions are grouped into a series according to their lower energy state (n’).
The Brackett series is the emission or absorption lines for H- atoms due to the electron jumps between the fourth energy level and other higher energy levels.
In the question, we are asked the number of possible spectral lines which may be emitted in the Brackett series in H-atom if the electrons in the ninth excited state return to the ground state.
Now for the Brackett series, the lowest energy level i.e. the ground state is n’=4 and the electron is in the ninth excited state.
We know that the first excited state will be the second energy level (as the electrons move from the first energy level to the second).
So, the ninth excited state will be the tenth energy level i.e. n = 10.
So, we have to calculate the number of possible spectral lines when the electron moves from n = 10 to n’ = 4.
So the possibilities are 10 $\to $ 4, 9 $\to $ 4, 8 $\to $ 4, 7 $\to $ 4, 6 $\to $ 4, 5$\to $ 4.
We can see from the above discussion that there are 6 possibilities.
Therefore, the correct answer is option [C] 6.
NOTE: For a hydrogen atom, the spectral lines are given special names.
- For n’=1 the series is called the Lyman series, which lie in the ultraviolet region,
- n’=2 is given the name Balmer series which has a few lines in the visible region,
- n’=3 is the Paschen series lying in the infrared region,
And there are others for n’= 4, 5 and 6 named as Brackett series, Pfund series and Humphrey series respectively.
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