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ATOMS

145526 The difference in the wavelength between the maximum and minimum of Balmer series [Use $R_{H} = 1 \times 10^{7} m^{- 1}$ ]

1

1600 \AA

2

3200 \AA ̊

3

4000 \AA

4

4800 \AA ̊

Explanation:

B Given that,
$R_{H} = 1 \times 10^{7} m^{- 1}$
Then minimum wavelength of Balmer series,
$n_{1} = 2 and n_{2} = \infty$
$\frac{1}{λ_{min}} = R_{H} (\frac{1}{2^{2}} - \frac{1}{\infty^{2}}) = \frac{R_{H}}{4}$
$λ_{min} = \frac{4}{R_{H}}$
For maximum wavelength of Balmer,
$n_{1} = 2 and n_{2} = 3$
Then, $\frac{1}{λ_{max}} = R_{H} (\frac{1}{2^{2}} - \frac{1}{3^{3}}) = R_{H} (\frac{1}{4} - \frac{1}{9}) = \frac{5 R_{H}}{36}$
$λ_{max} = \frac{36}{5 R_{H}}$
Then, difference $λ_{max} - λ_{min} = \frac{36}{5 R_{H}} - \frac{4}{R_{H}}$
$Δ λ = \frac{4}{R_{H}} (\frac{9}{5} - 1) = \frac{4}{R_{H}} \times \frac{4}{5}$
$Δ λ = \frac{16}{1 \times 10^{7} \times 5} = 3.2 \times 10^{- 7} m^{- 2}$
$Δ λ = 3200 \AA$

TSEAMCET 19.07.2022

ATOMS

145527 The ratio between shortest wavelength in Balmer series to shortest wavelength in Brackett series is

1

1 : 16

2

16 : 1

3

1 : 4

4

4 : 1

Explanation:

C For shortest wavelength of Balmer series,
$n_{1} = 2, n_{2} = \infty$
$\frac{1}{λ} = R (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) = R (\frac{1}{2^{2}} - \frac{1}{\infty^{2}})$
$\frac{1}{λ_{B}} = \frac{R}{4}$
And shortest wavelength of Bracket series,
$n_{1} = 4 and n_{2} = \infty$
$\frac{1}{λ_{Bracket}} = R (\frac{1}{4^{2}} - \frac{1}{\infty^{2}}) = \frac{R}{16}$
$\frac{1}{λ_{Bracket}} = \frac{R}{16}$
$\frac{λ_{Balmer}}{λ_{Bracket}} = \frac{R}{16} \times \frac{4}{R} = \frac{1}{4}$
$\frac{λ_{Balmer}}{λ_{Bracket}} = \frac{1}{4}$

TSEAMCET 30.07.2022

ATOMS

145529 The shortest wavelength in Balmer series of hydrogen atom spectrum is approximately equal to
$(use R_{H} = 1.097 \times 10^{7} m^{- 1})$

1

3646 \AA

2

912 \AA

3

364.6 \AA

4

91.2 \AA ̊

Explanation:

A Given that, $R_{H} = 1.097 \times 10^{7} m^{- 1}$
We know that,
$\frac{1}{λ} = R (\frac{1}{2^{2}} - \frac{1}{n^{2}})$
For shortest wavelength of Balmer series $n_{1} = 2, n_{2} = \infty$
$\frac{1}{λ} = R_{H} (\frac{1}{2^{2}} - \frac{1}{\infty^{2}}) = \frac{R_{H}}{4}$
$λ = \frac{4}{R_{H}} = \frac{4}{1.097 \times 10^{7}} = 364.6 \times 10^{- 10} m$
$λ = 3646 \AA$

TSEAMCET 20.07.2022

ATOMS

145532 The wavelength of the first spectral line of the Lyman series of hydrogen spectrum is

1

912 \AA

2

1215 \AA

3

1512 \AA

4

6563 \AA

Explanation:

B Wavelength of the first spectral line of the Lyman series of hydrogen
$\frac{1}{λ} = R (\frac{1}{1^{2}} - \frac{1}{2^{2}})$
$\frac{1}{λ} = 1.097 (1 - \frac{1}{4}) \times 10^{7}$
$\frac{1}{λ} = 1.097 (\frac{3}{4}) \times 10^{7}$
$λ = 1.21543 \times 10^{- 7}$
$λ = 1215 \AA$

AP EAMCET-20.08.2021

ATOMS

145526 The difference in the wavelength between the maximum and minimum of Balmer series [Use $R_{H} = 1 \times 10^{7} m^{- 1}$ ]

1

1600 \AA

2

3200 \AA ̊

3

4000 \AA

4

4800 \AA ̊

Explanation:

B Given that,
$R_{H} = 1 \times 10^{7} m^{- 1}$
Then minimum wavelength of Balmer series,
$n_{1} = 2 and n_{2} = \infty$
$\frac{1}{λ_{min}} = R_{H} (\frac{1}{2^{2}} - \frac{1}{\infty^{2}}) = \frac{R_{H}}{4}$
$λ_{min} = \frac{4}{R_{H}}$
For maximum wavelength of Balmer,
$n_{1} = 2 and n_{2} = 3$
Then, $\frac{1}{λ_{max}} = R_{H} (\frac{1}{2^{2}} - \frac{1}{3^{3}}) = R_{H} (\frac{1}{4} - \frac{1}{9}) = \frac{5 R_{H}}{36}$
$λ_{max} = \frac{36}{5 R_{H}}$
Then, difference $λ_{max} - λ_{min} = \frac{36}{5 R_{H}} - \frac{4}{R_{H}}$
$Δ λ = \frac{4}{R_{H}} (\frac{9}{5} - 1) = \frac{4}{R_{H}} \times \frac{4}{5}$
$Δ λ = \frac{16}{1 \times 10^{7} \times 5} = 3.2 \times 10^{- 7} m^{- 2}$
$Δ λ = 3200 \AA$

TSEAMCET 19.07.2022

ATOMS

145527 The ratio between shortest wavelength in Balmer series to shortest wavelength in Brackett series is

1

1 : 16

2

16 : 1

3

1 : 4

4

4 : 1

Explanation:

C For shortest wavelength of Balmer series,
$n_{1} = 2, n_{2} = \infty$
$\frac{1}{λ} = R (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) = R (\frac{1}{2^{2}} - \frac{1}{\infty^{2}})$
$\frac{1}{λ_{B}} = \frac{R}{4}$
And shortest wavelength of Bracket series,
$n_{1} = 4 and n_{2} = \infty$
$\frac{1}{λ_{Bracket}} = R (\frac{1}{4^{2}} - \frac{1}{\infty^{2}}) = \frac{R}{16}$
$\frac{1}{λ_{Bracket}} = \frac{R}{16}$
$\frac{λ_{Balmer}}{λ_{Bracket}} = \frac{R}{16} \times \frac{4}{R} = \frac{1}{4}$
$\frac{λ_{Balmer}}{λ_{Bracket}} = \frac{1}{4}$

TSEAMCET 30.07.2022

ATOMS

145529 The shortest wavelength in Balmer series of hydrogen atom spectrum is approximately equal to
$(use R_{H} = 1.097 \times 10^{7} m^{- 1})$

1

3646 \AA

2

912 \AA

3

364.6 \AA

4

91.2 \AA ̊

Explanation:

A Given that, $R_{H} = 1.097 \times 10^{7} m^{- 1}$
We know that,
$\frac{1}{λ} = R (\frac{1}{2^{2}} - \frac{1}{n^{2}})$
For shortest wavelength of Balmer series $n_{1} = 2, n_{2} = \infty$
$\frac{1}{λ} = R_{H} (\frac{1}{2^{2}} - \frac{1}{\infty^{2}}) = \frac{R_{H}}{4}$
$λ = \frac{4}{R_{H}} = \frac{4}{1.097 \times 10^{7}} = 364.6 \times 10^{- 10} m$
$λ = 3646 \AA$

TSEAMCET 20.07.2022

ATOMS

145532 The wavelength of the first spectral line of the Lyman series of hydrogen spectrum is

1

912 \AA

2

1215 \AA

3

1512 \AA

4

6563 \AA

Explanation:

B Wavelength of the first spectral line of the Lyman series of hydrogen
$\frac{1}{λ} = R (\frac{1}{1^{2}} - \frac{1}{2^{2}})$
$\frac{1}{λ} = 1.097 (1 - \frac{1}{4}) \times 10^{7}$
$\frac{1}{λ} = 1.097 (\frac{3}{4}) \times 10^{7}$
$λ = 1.21543 \times 10^{- 7}$
$λ = 1215 \AA$

AP EAMCET-20.08.2021

ATOMS

145526 The difference in the wavelength between the maximum and minimum of Balmer series [Use $R_{H} = 1 \times 10^{7} m^{- 1}$ ]

1

1600 \AA

2

3200 \AA ̊

3

4000 \AA

4

4800 \AA ̊

Explanation:

B Given that,
$R_{H} = 1 \times 10^{7} m^{- 1}$
Then minimum wavelength of Balmer series,
$n_{1} = 2 and n_{2} = \infty$
$\frac{1}{λ_{min}} = R_{H} (\frac{1}{2^{2}} - \frac{1}{\infty^{2}}) = \frac{R_{H}}{4}$
$λ_{min} = \frac{4}{R_{H}}$
For maximum wavelength of Balmer,
$n_{1} = 2 and n_{2} = 3$
Then, $\frac{1}{λ_{max}} = R_{H} (\frac{1}{2^{2}} - \frac{1}{3^{3}}) = R_{H} (\frac{1}{4} - \frac{1}{9}) = \frac{5 R_{H}}{36}$
$λ_{max} = \frac{36}{5 R_{H}}$
Then, difference $λ_{max} - λ_{min} = \frac{36}{5 R_{H}} - \frac{4}{R_{H}}$
$Δ λ = \frac{4}{R_{H}} (\frac{9}{5} - 1) = \frac{4}{R_{H}} \times \frac{4}{5}$
$Δ λ = \frac{16}{1 \times 10^{7} \times 5} = 3.2 \times 10^{- 7} m^{- 2}$
$Δ λ = 3200 \AA$

TSEAMCET 19.07.2022

ATOMS

145527 The ratio between shortest wavelength in Balmer series to shortest wavelength in Brackett series is

1

1 : 16

2

16 : 1

3

1 : 4

4

4 : 1

Explanation:

C For shortest wavelength of Balmer series,
$n_{1} = 2, n_{2} = \infty$
$\frac{1}{λ} = R (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) = R (\frac{1}{2^{2}} - \frac{1}{\infty^{2}})$
$\frac{1}{λ_{B}} = \frac{R}{4}$
And shortest wavelength of Bracket series,
$n_{1} = 4 and n_{2} = \infty$
$\frac{1}{λ_{Bracket}} = R (\frac{1}{4^{2}} - \frac{1}{\infty^{2}}) = \frac{R}{16}$
$\frac{1}{λ_{Bracket}} = \frac{R}{16}$
$\frac{λ_{Balmer}}{λ_{Bracket}} = \frac{R}{16} \times \frac{4}{R} = \frac{1}{4}$
$\frac{λ_{Balmer}}{λ_{Bracket}} = \frac{1}{4}$

TSEAMCET 30.07.2022

ATOMS

145529 The shortest wavelength in Balmer series of hydrogen atom spectrum is approximately equal to
$(use R_{H} = 1.097 \times 10^{7} m^{- 1})$

1

3646 \AA

2

912 \AA

3

364.6 \AA

4

91.2 \AA ̊

Explanation:

A Given that, $R_{H} = 1.097 \times 10^{7} m^{- 1}$
We know that,
$\frac{1}{λ} = R (\frac{1}{2^{2}} - \frac{1}{n^{2}})$
For shortest wavelength of Balmer series $n_{1} = 2, n_{2} = \infty$
$\frac{1}{λ} = R_{H} (\frac{1}{2^{2}} - \frac{1}{\infty^{2}}) = \frac{R_{H}}{4}$
$λ = \frac{4}{R_{H}} = \frac{4}{1.097 \times 10^{7}} = 364.6 \times 10^{- 10} m$
$λ = 3646 \AA$

TSEAMCET 20.07.2022

ATOMS

145532 The wavelength of the first spectral line of the Lyman series of hydrogen spectrum is

1

912 \AA

2

1215 \AA

3

1512 \AA

4

6563 \AA

Explanation:

B Wavelength of the first spectral line of the Lyman series of hydrogen
$\frac{1}{λ} = R (\frac{1}{1^{2}} - \frac{1}{2^{2}})$
$\frac{1}{λ} = 1.097 (1 - \frac{1}{4}) \times 10^{7}$
$\frac{1}{λ} = 1.097 (\frac{3}{4}) \times 10^{7}$
$λ = 1.21543 \times 10^{- 7}$
$λ = 1215 \AA$

AP EAMCET-20.08.2021

ATOMS

145526 The difference in the wavelength between the maximum and minimum of Balmer series [Use $R_{H} = 1 \times 10^{7} m^{- 1}$ ]

1

1600 \AA

2

3200 \AA ̊

3

4000 \AA

4

4800 \AA ̊

Explanation:

B Given that,
$R_{H} = 1 \times 10^{7} m^{- 1}$
Then minimum wavelength of Balmer series,
$n_{1} = 2 and n_{2} = \infty$
$\frac{1}{λ_{min}} = R_{H} (\frac{1}{2^{2}} - \frac{1}{\infty^{2}}) = \frac{R_{H}}{4}$
$λ_{min} = \frac{4}{R_{H}}$
For maximum wavelength of Balmer,
$n_{1} = 2 and n_{2} = 3$
Then, $\frac{1}{λ_{max}} = R_{H} (\frac{1}{2^{2}} - \frac{1}{3^{3}}) = R_{H} (\frac{1}{4} - \frac{1}{9}) = \frac{5 R_{H}}{36}$
$λ_{max} = \frac{36}{5 R_{H}}$
Then, difference $λ_{max} - λ_{min} = \frac{36}{5 R_{H}} - \frac{4}{R_{H}}$
$Δ λ = \frac{4}{R_{H}} (\frac{9}{5} - 1) = \frac{4}{R_{H}} \times \frac{4}{5}$
$Δ λ = \frac{16}{1 \times 10^{7} \times 5} = 3.2 \times 10^{- 7} m^{- 2}$
$Δ λ = 3200 \AA$

TSEAMCET 19.07.2022

ATOMS

145527 The ratio between shortest wavelength in Balmer series to shortest wavelength in Brackett series is

1

1 : 16

2

16 : 1

3

1 : 4

4

4 : 1

Explanation:

C For shortest wavelength of Balmer series,
$n_{1} = 2, n_{2} = \infty$
$\frac{1}{λ} = R (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) = R (\frac{1}{2^{2}} - \frac{1}{\infty^{2}})$
$\frac{1}{λ_{B}} = \frac{R}{4}$
And shortest wavelength of Bracket series,
$n_{1} = 4 and n_{2} = \infty$
$\frac{1}{λ_{Bracket}} = R (\frac{1}{4^{2}} - \frac{1}{\infty^{2}}) = \frac{R}{16}$
$\frac{1}{λ_{Bracket}} = \frac{R}{16}$
$\frac{λ_{Balmer}}{λ_{Bracket}} = \frac{R}{16} \times \frac{4}{R} = \frac{1}{4}$
$\frac{λ_{Balmer}}{λ_{Bracket}} = \frac{1}{4}$

TSEAMCET 30.07.2022

ATOMS

145529 The shortest wavelength in Balmer series of hydrogen atom spectrum is approximately equal to
$(use R_{H} = 1.097 \times 10^{7} m^{- 1})$

1

3646 \AA

2

912 \AA

3

364.6 \AA

4

91.2 \AA ̊

Explanation:

A Given that, $R_{H} = 1.097 \times 10^{7} m^{- 1}$
We know that,
$\frac{1}{λ} = R (\frac{1}{2^{2}} - \frac{1}{n^{2}})$
For shortest wavelength of Balmer series $n_{1} = 2, n_{2} = \infty$
$\frac{1}{λ} = R_{H} (\frac{1}{2^{2}} - \frac{1}{\infty^{2}}) = \frac{R_{H}}{4}$
$λ = \frac{4}{R_{H}} = \frac{4}{1.097 \times 10^{7}} = 364.6 \times 10^{- 10} m$
$λ = 3646 \AA$

TSEAMCET 20.07.2022

ATOMS

145532 The wavelength of the first spectral line of the Lyman series of hydrogen spectrum is

1

912 \AA

2

1215 \AA

3

1512 \AA

4

6563 \AA

Explanation:

B Wavelength of the first spectral line of the Lyman series of hydrogen
$\frac{1}{λ} = R (\frac{1}{1^{2}} - \frac{1}{2^{2}})$
$\frac{1}{λ} = 1.097 (1 - \frac{1}{4}) \times 10^{7}$
$\frac{1}{λ} = 1.097 (\frac{3}{4}) \times 10^{7}$
$λ = 1.21543 \times 10^{- 7}$
$λ = 1215 \AA$

AP EAMCET-20.08.2021

Question Bank Series

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