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CHXI02:STRUCTURE OF ATOM

307591 If the shortest wavelength of the spectral line of H-atom in the Lyman series is X, then the longest wavelength of the line in Balmer series of $L i^{2 +}$ is

1

9 x

2

\frac{x}{9}

3

\frac{5 x}{4}

4

\frac{4 x}{5}

Explanation:

$\frac{1}{λ} = R_{H} Z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
To calculate shortest wavelength take $n_{2} = \infty$ and longest wavelength take nearest value of $n_{2}$ .
For H-atom,
$\frac{1}{λ_{s h o r t e s t}}, n_{2} = \infty, Z = 1, n_{1} = 1$
$∴ \frac{1}{x} = R_{H}$ (Lyman series)
For $\frac{1}{λ_{l o n g e s t}} f o r L i^{2 +}, Z = 3, n_{1} = 2, n_{2} = 3$
(Balmer series)
$\frac{1}{λ_{l o n g e s t}} = \frac{1}{x} \times 3^{2} (\frac{1}{2^{2}} - \frac{1}{3^{2}}) = \frac{5}{4 x}$
$∴ λ_{l o n g e s t} = \frac{4 x}{5}$

CHXI02:STRUCTURE OF ATOM

307592 For hydrogen atom, the shortest wavelength of Lyman series is $λ_{L}$ $A^{\circ}$ while the shortest wavelength of Paschen series is $λ_{P}$ $A^{\circ}$ . The value of $\frac{λ_{P}}{λ_{L}}$ is

1 9

2

\frac{1}{9}

3

\infty

4 1

Explanation:

For shortest wavelength, $n_{2} = \infty$
$∴$ For Lyman series, $\frac{1}{λ_{L}} = R_{H} [\frac{1}{1^{2}} - \frac{1}{\infty^{2}}] = R_{H} [1]$
For Paschen series, $\frac{1}{λ_{p}} = R_{H} [\frac{1}{(3)^{2}} - \frac{1}{\infty^{2}}]$
$= R_{H} [\frac{1}{9}]$
$∴ λ_{L} = \frac{1}{R_{H} [1]} = \frac{1}{R_{H}}$ and $λ_{p} = \frac{1}{R_{H} [\frac{1}{9}]} = \frac{9}{R_{H}}$
$∴ \frac{λ_{p}}{λ_{ℓ}} = \frac{9/ R_{H}}{1/ R_{H}} = 9$

CHXI02:STRUCTURE OF ATOM

307593 The wavelength of the third line of the Balmer series for a hydrogen atom is

1

\frac{21}{100 R_{H}}

2

\frac{100}{21 R_{H}}

3

\frac{21 R_{H}}{100}

4

\frac{100 R_{H}}{21}

Explanation:

The third line of Balmer series possesses $n_{1} = 2$ and $n_{2} = 5$
By using Rydberg equation,
$\frac{1}{λ} = R_{H} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) = R_{H} (\frac{1}{2^{2}} - \frac{1}{5^{2}}) = R_{H} (\frac{21}{100})$
$λ = \frac{100}{21 R_{H}}$

CHXI02:STRUCTURE OF ATOM

307594 Determine ratio of wavelength of first line & third line of Balmer series in H-Spectrum

1

1

2

1.512

3

2

4

2 .512

Explanation:

Transition for $1^{s t}$ line of Balmer series, $3 \to 2$
$\frac{1}{λ} = R (\frac{1}{2^{2}} - \frac{1}{3^{2}}) = R (\frac{1}{4} - \frac{1}{9}) = \frac{5 R}{36}$
$λ = \frac{36}{5 R}$
Transition for line of Balmer series, $5 \to 2$
$\frac{1}{λ} = R (\frac{1}{2^{2}} - \frac{1}{5^{2}}) = R (\frac{1}{4} - \frac{1}{25}) = \frac{21 R}{100}$
$λ = \frac{100}{21 R}$
Ratio of wavelength is $= \frac{\frac{36}{5 R}}{\frac{100}{21 R}} = 1.512$

JEE - 2021

CHXI02:STRUCTURE OF ATOM

307591 If the shortest wavelength of the spectral line of H-atom in the Lyman series is X, then the longest wavelength of the line in Balmer series of $L i^{2 +}$ is

1

9 x

2

\frac{x}{9}

3

\frac{5 x}{4}

4

\frac{4 x}{5}

Explanation:

$\frac{1}{λ} = R_{H} Z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
To calculate shortest wavelength take $n_{2} = \infty$ and longest wavelength take nearest value of $n_{2}$ .
For H-atom,
$\frac{1}{λ_{s h o r t e s t}}, n_{2} = \infty, Z = 1, n_{1} = 1$
$∴ \frac{1}{x} = R_{H}$ (Lyman series)
For $\frac{1}{λ_{l o n g e s t}} f o r L i^{2 +}, Z = 3, n_{1} = 2, n_{2} = 3$
(Balmer series)
$\frac{1}{λ_{l o n g e s t}} = \frac{1}{x} \times 3^{2} (\frac{1}{2^{2}} - \frac{1}{3^{2}}) = \frac{5}{4 x}$
$∴ λ_{l o n g e s t} = \frac{4 x}{5}$

CHXI02:STRUCTURE OF ATOM

307592 For hydrogen atom, the shortest wavelength of Lyman series is $λ_{L}$ $A^{\circ}$ while the shortest wavelength of Paschen series is $λ_{P}$ $A^{\circ}$ . The value of $\frac{λ_{P}}{λ_{L}}$ is

1 9

2

\frac{1}{9}

3

\infty

4 1

Explanation:

For shortest wavelength, $n_{2} = \infty$
$∴$ For Lyman series, $\frac{1}{λ_{L}} = R_{H} [\frac{1}{1^{2}} - \frac{1}{\infty^{2}}] = R_{H} [1]$
For Paschen series, $\frac{1}{λ_{p}} = R_{H} [\frac{1}{(3)^{2}} - \frac{1}{\infty^{2}}]$
$= R_{H} [\frac{1}{9}]$
$∴ λ_{L} = \frac{1}{R_{H} [1]} = \frac{1}{R_{H}}$ and $λ_{p} = \frac{1}{R_{H} [\frac{1}{9}]} = \frac{9}{R_{H}}$
$∴ \frac{λ_{p}}{λ_{ℓ}} = \frac{9/ R_{H}}{1/ R_{H}} = 9$

CHXI02:STRUCTURE OF ATOM

307593 The wavelength of the third line of the Balmer series for a hydrogen atom is

1

\frac{21}{100 R_{H}}

2

\frac{100}{21 R_{H}}

3

\frac{21 R_{H}}{100}

4

\frac{100 R_{H}}{21}

Explanation:

The third line of Balmer series possesses $n_{1} = 2$ and $n_{2} = 5$
By using Rydberg equation,
$\frac{1}{λ} = R_{H} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) = R_{H} (\frac{1}{2^{2}} - \frac{1}{5^{2}}) = R_{H} (\frac{21}{100})$
$λ = \frac{100}{21 R_{H}}$

CHXI02:STRUCTURE OF ATOM

307594 Determine ratio of wavelength of first line & third line of Balmer series in H-Spectrum

1

1

2

1.512

3

2

4

2 .512

Explanation:

Transition for $1^{s t}$ line of Balmer series, $3 \to 2$
$\frac{1}{λ} = R (\frac{1}{2^{2}} - \frac{1}{3^{2}}) = R (\frac{1}{4} - \frac{1}{9}) = \frac{5 R}{36}$
$λ = \frac{36}{5 R}$
Transition for line of Balmer series, $5 \to 2$
$\frac{1}{λ} = R (\frac{1}{2^{2}} - \frac{1}{5^{2}}) = R (\frac{1}{4} - \frac{1}{25}) = \frac{21 R}{100}$
$λ = \frac{100}{21 R}$
Ratio of wavelength is $= \frac{\frac{36}{5 R}}{\frac{100}{21 R}} = 1.512$

JEE - 2021

CHXI02:STRUCTURE OF ATOM

307591 If the shortest wavelength of the spectral line of H-atom in the Lyman series is X, then the longest wavelength of the line in Balmer series of $L i^{2 +}$ is

1

9 x

2

\frac{x}{9}

3

\frac{5 x}{4}

4

\frac{4 x}{5}

Explanation:

$\frac{1}{λ} = R_{H} Z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
To calculate shortest wavelength take $n_{2} = \infty$ and longest wavelength take nearest value of $n_{2}$ .
For H-atom,
$\frac{1}{λ_{s h o r t e s t}}, n_{2} = \infty, Z = 1, n_{1} = 1$
$∴ \frac{1}{x} = R_{H}$ (Lyman series)
For $\frac{1}{λ_{l o n g e s t}} f o r L i^{2 +}, Z = 3, n_{1} = 2, n_{2} = 3$
(Balmer series)
$\frac{1}{λ_{l o n g e s t}} = \frac{1}{x} \times 3^{2} (\frac{1}{2^{2}} - \frac{1}{3^{2}}) = \frac{5}{4 x}$
$∴ λ_{l o n g e s t} = \frac{4 x}{5}$

CHXI02:STRUCTURE OF ATOM

307592 For hydrogen atom, the shortest wavelength of Lyman series is $λ_{L}$ $A^{\circ}$ while the shortest wavelength of Paschen series is $λ_{P}$ $A^{\circ}$ . The value of $\frac{λ_{P}}{λ_{L}}$ is

1 9

2

\frac{1}{9}

3

\infty

4 1

Explanation:

For shortest wavelength, $n_{2} = \infty$
$∴$ For Lyman series, $\frac{1}{λ_{L}} = R_{H} [\frac{1}{1^{2}} - \frac{1}{\infty^{2}}] = R_{H} [1]$
For Paschen series, $\frac{1}{λ_{p}} = R_{H} [\frac{1}{(3)^{2}} - \frac{1}{\infty^{2}}]$
$= R_{H} [\frac{1}{9}]$
$∴ λ_{L} = \frac{1}{R_{H} [1]} = \frac{1}{R_{H}}$ and $λ_{p} = \frac{1}{R_{H} [\frac{1}{9}]} = \frac{9}{R_{H}}$
$∴ \frac{λ_{p}}{λ_{ℓ}} = \frac{9/ R_{H}}{1/ R_{H}} = 9$

CHXI02:STRUCTURE OF ATOM

307593 The wavelength of the third line of the Balmer series for a hydrogen atom is

1

\frac{21}{100 R_{H}}

2

\frac{100}{21 R_{H}}

3

\frac{21 R_{H}}{100}

4

\frac{100 R_{H}}{21}

Explanation:

The third line of Balmer series possesses $n_{1} = 2$ and $n_{2} = 5$
By using Rydberg equation,
$\frac{1}{λ} = R_{H} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) = R_{H} (\frac{1}{2^{2}} - \frac{1}{5^{2}}) = R_{H} (\frac{21}{100})$
$λ = \frac{100}{21 R_{H}}$

CHXI02:STRUCTURE OF ATOM

307594 Determine ratio of wavelength of first line & third line of Balmer series in H-Spectrum

1

1

2

1.512

3

2

4

2 .512

Explanation:

Transition for $1^{s t}$ line of Balmer series, $3 \to 2$
$\frac{1}{λ} = R (\frac{1}{2^{2}} - \frac{1}{3^{2}}) = R (\frac{1}{4} - \frac{1}{9}) = \frac{5 R}{36}$
$λ = \frac{36}{5 R}$
Transition for line of Balmer series, $5 \to 2$
$\frac{1}{λ} = R (\frac{1}{2^{2}} - \frac{1}{5^{2}}) = R (\frac{1}{4} - \frac{1}{25}) = \frac{21 R}{100}$
$λ = \frac{100}{21 R}$
Ratio of wavelength is $= \frac{\frac{36}{5 R}}{\frac{100}{21 R}} = 1.512$

JEE - 2021

CHXI02:STRUCTURE OF ATOM

307591 If the shortest wavelength of the spectral line of H-atom in the Lyman series is X, then the longest wavelength of the line in Balmer series of $L i^{2 +}$ is

1

9 x

2

\frac{x}{9}

3

\frac{5 x}{4}

4

\frac{4 x}{5}

Explanation:

$\frac{1}{λ} = R_{H} Z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
To calculate shortest wavelength take $n_{2} = \infty$ and longest wavelength take nearest value of $n_{2}$ .
For H-atom,
$\frac{1}{λ_{s h o r t e s t}}, n_{2} = \infty, Z = 1, n_{1} = 1$
$∴ \frac{1}{x} = R_{H}$ (Lyman series)
For $\frac{1}{λ_{l o n g e s t}} f o r L i^{2 +}, Z = 3, n_{1} = 2, n_{2} = 3$
(Balmer series)
$\frac{1}{λ_{l o n g e s t}} = \frac{1}{x} \times 3^{2} (\frac{1}{2^{2}} - \frac{1}{3^{2}}) = \frac{5}{4 x}$
$∴ λ_{l o n g e s t} = \frac{4 x}{5}$

CHXI02:STRUCTURE OF ATOM

307592 For hydrogen atom, the shortest wavelength of Lyman series is $λ_{L}$ $A^{\circ}$ while the shortest wavelength of Paschen series is $λ_{P}$ $A^{\circ}$ . The value of $\frac{λ_{P}}{λ_{L}}$ is

1 9

2

\frac{1}{9}

3

\infty

4 1

Explanation:

For shortest wavelength, $n_{2} = \infty$
$∴$ For Lyman series, $\frac{1}{λ_{L}} = R_{H} [\frac{1}{1^{2}} - \frac{1}{\infty^{2}}] = R_{H} [1]$
For Paschen series, $\frac{1}{λ_{p}} = R_{H} [\frac{1}{(3)^{2}} - \frac{1}{\infty^{2}}]$
$= R_{H} [\frac{1}{9}]$
$∴ λ_{L} = \frac{1}{R_{H} [1]} = \frac{1}{R_{H}}$ and $λ_{p} = \frac{1}{R_{H} [\frac{1}{9}]} = \frac{9}{R_{H}}$
$∴ \frac{λ_{p}}{λ_{ℓ}} = \frac{9/ R_{H}}{1/ R_{H}} = 9$

CHXI02:STRUCTURE OF ATOM

307593 The wavelength of the third line of the Balmer series for a hydrogen atom is

1

\frac{21}{100 R_{H}}

2

\frac{100}{21 R_{H}}

3

\frac{21 R_{H}}{100}

4

\frac{100 R_{H}}{21}

Explanation:

The third line of Balmer series possesses $n_{1} = 2$ and $n_{2} = 5$
By using Rydberg equation,
$\frac{1}{λ} = R_{H} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) = R_{H} (\frac{1}{2^{2}} - \frac{1}{5^{2}}) = R_{H} (\frac{21}{100})$
$λ = \frac{100}{21 R_{H}}$

CHXI02:STRUCTURE OF ATOM

307594 Determine ratio of wavelength of first line & third line of Balmer series in H-Spectrum

1

1

2

1.512

3

2

4

2 .512

Explanation:

Transition for $1^{s t}$ line of Balmer series, $3 \to 2$
$\frac{1}{λ} = R (\frac{1}{2^{2}} - \frac{1}{3^{2}}) = R (\frac{1}{4} - \frac{1}{9}) = \frac{5 R}{36}$
$λ = \frac{36}{5 R}$
Transition for line of Balmer series, $5 \to 2$
$\frac{1}{λ} = R (\frac{1}{2^{2}} - \frac{1}{5^{2}}) = R (\frac{1}{4} - \frac{1}{25}) = \frac{21 R}{100}$
$λ = \frac{100}{21 R}$
Ratio of wavelength is $= \frac{\frac{36}{5 R}}{\frac{100}{21 R}} = 1.512$

JEE - 2021