Question

The shortest wavelengths emitted in the hydrogen spectrum corresponding to different spectral series are:
(A) Pfund series
(B) Balmer series
(C) Brackett series
(D) Lyman series.
The wavelengths arranged correctly in decreasing order are_______.
Fill in the blank with the correct answer from the options given below.

• A. (A), (B), (C), (D)
• B. (A), (C), (B), (D)
• C. (B), (A), (D), (C)
• D. (A), (C), (D), (B)

Answer 1

The Correct Option is B

The hydrogen spectrum is characterized by different spectral series, each corresponding to electron transitions between energy levels. The shortest wavelength in each series occurs when the transition is from the higher energy levels to the lowest level of that series. These series in the order of the smallest to largest shortest wavelengths are:

1. Lyman series (ultraviolet region): Transitions are to n=1. The shortest wavelength corresponds to the transition from n=∞ to n=1.
2. Balmer series (visible region): Transitions are to n=2. The shortest wavelength corresponds to the transition from n=∞ to n=2.
3. Paschen series (infrared region): Transitions are to n=3.
4. Brackett series (infrared region): Transitions are to n=4.
5. Pfund series (far infrared region): Transitions are to n=5. The shortest wavelength corresponds to the transition from n=∞ to n=5.

Therefore, the arrangement of these series in the order of decreasing wavelength (from longest to shortest) is:
(A) Pfund series
(C) Brackett series
(B) Balmer series
(D) Lyman series
Hence, the correct order is: (A), (C), (B), (D)

Answer 2

To determine the shortest wavelengths emitted in the hydrogen spectrum for different spectral series, we need to understand the transitions involved.

The shortest wavelength in a spectral series corresponds to the transition from n = ∞ to the lowest energy level of that series.

The spectral series are defined as follows:

Lyman series (D): Transitions to n = 1 (ultraviolet region)

Balmer series (B): Transitions to n = 2 (visible region)

Paschen series: Transitions to n = 3 (infrared region)

Brackett series (C): Transitions to n = 4 (infrared region)

Pfund series (A): Transitions to n = 5 (far-infrared region)

The energy difference between the levels determines the wavelength of the emitted photon. The larger the energy difference, the shorter the wavelength.

The energy levels are given by:

E_n = -13.6 eV / n²

The transition from n = ∞ to the lowest level (n = 1, 2, 3, 4, 5) results in the shortest wavelength for each series.

The energy difference is greatest for the Lyman series (n = 1), followed by the Balmer series (n = 2), then the Brackett series (n = 4), and then the Pfund series (n = 5).

Therefore, the shortest wavelengths in decreasing order are:

Lyman > Balmer > Brackett > Pfund

Which corresponds to:

(D), (B), (C), (A)

However, the question asks for the order of (A),(B),(C),(D) which means we need to rearrange the order to fit the given options. The correct order from largest to smallest n value for the final state is therefore from (A) Pfund to (D) Lyman.

The order of decreasing wavelength (and therefore increasing energy) is:

Pfund (A) > Brackett (C) > Balmer (B) > Lyman (D)

Therefore, the order of decreasing wavelength is (A), (C), (B), (D).

The correct answer is:

Option 2: (A), (C), (B), (D)