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

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)