The wavelength of the first member of the Balmer series is 6563 A o . Find the wavelengths of the second and the last member of the series.

For the first line of the Balmer series n 21 = 2 and n 22 = 3 The wave number ν of the associated wave will be: ν 1 = R H [ 1 n 2 21 − 1 n 2 22 ] cm − 1 Where, Rydberg Constant, R H = 109677 cm - 1 Substitute the values in the equation, = 109677 [ 1 2 2 − 1 3 2 ] cm − 1 = 109677 × 5 36 cm − 1 ∵ λ 1 = 1 ν 1 = 36 548385 ------(i) For the second line of the Balmer series n 21 = 2 and n 23 = 4 The wave number ν of the associated wave will be: ν 2 = R H [ 1 n 2 21 − 1 n 2 23 ] cm − 1 Where, Rydberg Constant, R H = 109677 cm - 1 Substitute the values in the equation, = 109677 [ 1 2 2 − 1 4 2 ] cm − 1 = 109677 × 3 16 cm − 1 ∵ λ 2 = 1 ν 2 = 16 329031 --------(ii) For the last line of the Balmer series the transition will take place between n 1 = 2 and n 2 = ∞ The wave number ν of the associated wave will be: ν ∞ = R H [ 1 n 2 1 − 1 n 2 2 ] cm − 1 Where, Rydberg Constant, R H = 109677 cm - 1 Substitute the values in the equation, = 109677 [ 1 2 2 − 1 ∞ 2 ] cm − 1 = 109677 × 1 4 cm − 1 ∵ λ ∞ = 1 ν = 4 109677 -----(iii) We are given λ 1 = 6563 A ° Divide equation (ii) by equation (i): λ 2 λ 1 = 20 27 λ 2 = 20 27 × 6563 = 4861 . 5 A ° Divide equation (iii) by equation (i): λ ∞ λ 1 = 20 36 λ ∞ = 20 36 × 6563 = 3646 . 1 A °

The wavelength of the first member of Balmer series is 6563 Å . Calculate the wavelength of second member of Lyman series.

According to Rydberg formula, 1 λ = R 1 n 1 2 - 1 n 2 2 where R is the Rydberg constant. For the first member of Balmer series, n 1 = 2 , n 2 = 3 ∴ 1 λ = R 1 2 2 - 1 3 2 1 λ = 5 R 36 R = 36 5 λ - - - ( i ) For the second member of Lyman series, n 1 = 1 , n 2 = 3 ∴ 1 λ ' = R 1 1 2 - 1 3 2 λ ' = 9 8 R Putting the value of R from ( i ) , we get, λ ' = 9 8 5 λ 36 λ ' = 5 32 λ Given: λ = 6563 Å λ ' = 5 32 6563 ⇒ 1025 . 5 Å

Describe Rutherford atom model. What are the draw backs of this model?

In 1911 , Rutherford conducted an experiment in which he bombarded alpha particles on a thin gold foil. In that experiment, he observed that: (i) Most of the alpha particles passed straight through the gold foil. (ii) Very few particles scattered by more than 1 ° . (iii) About 1 in 8000 particles deflected by more than 90 ° . Based on these observations, he came up with a new model of atom. According to his model of the atom: (i) Most of the atom was empty space with its entire positive charge concentrated in a very small space inside the atom. This space was called the nucleus. (ii) The size of the nucleus is very small as compared to the size of the atom. (iii) The nucleus is surrounded by negatively charged particles called electrons. The negative charge of electrons is equal to the positive charge of the nucleus. Therefore, an atom is electrically neutral. (iv) Electrons revolve around the nucleus in circular orbits. Drawbacks of Rutherford's model of atom: (i) According to Rutherford's model of atom, electrons revolve around the nucleus in a circular path and are constantly under centripetal force. As a result, the electrons must undergo acceleration. Charged particles undergoing acceleration radiates energy, thus, electrons loses energy continuously and eventually falls into the nucleus. Thus, it failed to explain the stability of the atom. (ii) According to Rutherford's model, the atoms should emit continuous spectrum instead of line spectrum. Thus, it failed to explain the existence of spectral lines in the hydrogen spectrum.

What are the limitations of Bohr's theory of hydrogen atom?

Limitations of Bohr's theory: (i) Bohr's theory is limited to simple atoms like hydrogen for which Z = 1 . It fails for atoms of elements for which Z > 1 . (ii) It failed to explain the Zeeman effect i.e when the spectral line gets split into various components in the presence of magnetic field. (iii) It failed to explain the Stark effect i.e when the spectral line gets split into almost negligible lines in the presence of electric field. (iv) It violates Heisenberg uncertainty principle.

E M B I B E

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 13 - Atoms>Exercise>Q 22

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IMPORTANT

Earn 100

The second member of Lyman series in hydrogen spectrum has wavelength $5400 Å$ . Find the wavelength of first member.

Important Questions on Atoms

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IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 13 - Atoms>Exercise>Q 23

Using the Rydberg formula, calculate the wavelength of the first four spectral lines in the Balmer series of the hydrogen spectrum.

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IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 13 - Atoms>Exercise>Q 24

Explain the different types of spectral series.

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IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 13 - Atoms>Exercise>Q 25

Discuss the spectral series of hydrogen atom.

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IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 13 - Atoms>Exercise>Q 26

The wavelength of the first member of the Balmer series is

6563 \overset{o}{A}

. Find the wavelengths of the second and the last member of the series.

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12th Telangana Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 13 - Atoms>Exercise>Q 27

What is the physical meaning of 'negative energy of an electron'?

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Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 13 - Atoms>Exercise>Q 28

Name a physical quantity whose dimensions are the same as those of angular momentum.

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Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 13 - Atoms>Exercise>Q 29

What do you understand by the phrase 'ground state atom'?

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IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 13 - Atoms>Exercise>Q 30

Derive an expression for potential and kinetic energy of an electron in any orbit of a hydrogen atom according to Bohr's atomic model. How does P.E change with increasing n?