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In H-atom, electron transits from 6th orbit to 2nd orbit in multi step. Then total spectral lines (without Balmer series) will be :- 1) 6 2)10 3)4 4) 0 Ans~6 How?
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In H-atom, electron transits from 6th orbit to 2nd orbit in multi step...
No.of spectral lines = Σ delta n.=Σ(6-2).= Σ 4..=4+3+2+1.here ,as electron jumps from 6 to 2 in Lyman series lines are not observed....4-balmer,3-paschen,2-bracket,1-pfund.without balmer series, no. of spectral lines is =3+2+1=6...
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In H-atom, electron transits from 6th orbit to 2nd orbit in multi step...
Explanation:
When an electron transits from a higher energy level to a lower energy level, energy is released in the form of electromagnetic radiation. The frequency of this radiation is related to the energy difference between the two levels by the equation E = hf, where E is the energy difference, h is Planck's constant, and f is the frequency of the radiation.
Calculating the Spectral Lines:
In the case of an electron transitioning from the 6th orbit to the 2nd orbit in a hydrogen atom, there will be several steps involved. Each step will result in the emission of a photon with a specific frequency, which will correspond to a specific spectral line.
To calculate the total number of spectral lines that will be produced in this process, we need to consider all the possible transitions that can occur between the energy levels involved. There are several ways to do this, but one common method is to use the formula:
N = (n^2 - m^2) / 2
where N is the number of spectral lines, n is the initial energy level, and m is the final energy level.
Using this formula, we can calculate the number of spectral lines produced by each step in the transition process:
- From n=6 to n=5: N = (6^2 - 5^2) / 2 = 5
- From n=5 to n=4: N = (5^2 - 4^2) / 2 = 6
- From n=4 to n=3: N = (4^2 - 3^2) / 2 = 5
- From n=3 to n=2: N = (3^2 - 2^2) / 2 = 4
Therefore, the total number of spectral lines produced by the electron transitioning from the 6th orbit to the 2nd orbit is:
5 + 6 + 5 + 4 = 20
However, we need to exclude the spectral lines that are part of the Balmer series, which are produced by transitions to the 2nd energy level (n=2). These include the famous H-alpha (656.3 nm), H-beta (486.1 nm), H-gamma (434.0 nm), and H-delta (410.2 nm) lines. Since we are excluding these lines, the total number of spectral lines produced by the transition from the 6th orbit to the 2nd orbit is:
20 - 4 = 16
Therefore, the answer to the question is 16 spectral lines, not 6.
When an electron transits from a higher energy level to a lower energy level, energy is released in the form of electromagnetic radiation. The frequency of this radiation is related to the energy difference between the two levels by the equation E = hf, where E is the energy difference, h is Planck's constant, and f is the frequency of the radiation.
Calculating the Spectral Lines:
In the case of an electron transitioning from the 6th orbit to the 2nd orbit in a hydrogen atom, there will be several steps involved. Each step will result in the emission of a photon with a specific frequency, which will correspond to a specific spectral line.
To calculate the total number of spectral lines that will be produced in this process, we need to consider all the possible transitions that can occur between the energy levels involved. There are several ways to do this, but one common method is to use the formula:
N = (n^2 - m^2) / 2
where N is the number of spectral lines, n is the initial energy level, and m is the final energy level.
Using this formula, we can calculate the number of spectral lines produced by each step in the transition process:
- From n=6 to n=5: N = (6^2 - 5^2) / 2 = 5
- From n=5 to n=4: N = (5^2 - 4^2) / 2 = 6
- From n=4 to n=3: N = (4^2 - 3^2) / 2 = 5
- From n=3 to n=2: N = (3^2 - 2^2) / 2 = 4
Therefore, the total number of spectral lines produced by the electron transitioning from the 6th orbit to the 2nd orbit is:
5 + 6 + 5 + 4 = 20
However, we need to exclude the spectral lines that are part of the Balmer series, which are produced by transitions to the 2nd energy level (n=2). These include the famous H-alpha (656.3 nm), H-beta (486.1 nm), H-gamma (434.0 nm), and H-delta (410.2 nm) lines. Since we are excluding these lines, the total number of spectral lines produced by the transition from the 6th orbit to the 2nd orbit is:
20 - 4 = 16
Therefore, the answer to the question is 16 spectral lines, not 6.
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In H-atom, electron transits from 6th orbit to 2nd orbit in multi step. Then total spectral lines (without Balmer series) will be :- 1) 6 2)10 3)4 4) 0 Ans~6 How?
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In H-atom, electron transits from 6th orbit to 2nd orbit in multi step. Then total spectral lines (without Balmer series) will be :- 1) 6 2)10 3)4 4) 0 Ans~6 How? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about In H-atom, electron transits from 6th orbit to 2nd orbit in multi step. Then total spectral lines (without Balmer series) will be :- 1) 6 2)10 3)4 4) 0 Ans~6 How? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In H-atom, electron transits from 6th orbit to 2nd orbit in multi step. Then total spectral lines (without Balmer series) will be :- 1) 6 2)10 3)4 4) 0 Ans~6 How?.
In H-atom, electron transits from 6th orbit to 2nd orbit in multi step. Then total spectral lines (without Balmer series) will be :- 1) 6 2)10 3)4 4) 0 Ans~6 How? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about In H-atom, electron transits from 6th orbit to 2nd orbit in multi step. Then total spectral lines (without Balmer series) will be :- 1) 6 2)10 3)4 4) 0 Ans~6 How? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In H-atom, electron transits from 6th orbit to 2nd orbit in multi step. Then total spectral lines (without Balmer series) will be :- 1) 6 2)10 3)4 4) 0 Ans~6 How?.
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