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3 October, 00:11

The Balmer series, named after Johann Balmer, is a portion of the hydrogen emission spectrum produced from the transitions between the n=2 energy level and higher energy levels. Part A Match each emission wavelength of the Balmer series to the corresponding transition. Drag each item to the appropriate bin. n=7 to n=2 n=6 to n=2 n=5 to n=2 n=4 to n=2 n=3 to n=2 match 410nm 397nm 486nm 656nm 434nm

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  1. 3 October, 01:52
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    The wavelength of the balmer series is calculated using the following steps;

    - Find the Principle Quantum Number for the Transition

    - Calculate the Term in Brackets

    - Multiply by the Rydberg Constant

    - Find the Wavelength

    The Balmer series in a hydrogen atom relates the possible electron transitions down to the n = 2 position to the wavelength of the emission that scientists observe.

    The λ symbol represents the wavelength, and RH is the Rydberg constant for hydrogen, with RH = 1.0968 * 107 m-1

    n=7 to n=2

    - The principal quantum numbers are 2 and 7.

    - (1/2²) - (1 / n²₂)

    For n₂ = 7, you get:

    (1/2²) - (1 / n²₂) = (1/2²) - (1 / 7²)

    = (1/4) - (1/49)

    = 0.2230

    - RH = 1.0968 * 107 m-1, to find a value for 1/λ. The formula and the example calculation gives:

    1/λ = RH [ (1/2²) - (1 / n²₂) ]

    = 1.0968 * 107 m-1 * 0.2230

    = 2445864 m-1

    - λ = 1 / 2445864 m-1

    = 4.08 * 10-7 m

    = 408 nanometers

    ≈ 410nm

    n=6 to n=2

    - The principal quantum numbers are 2 and 6.

    - (1/2²) - (1 / n²₂)

    For n₂ = 6, you get:

    (1/2²) - (1 / n²₂) = (1/2²) - (1 / 6²)

    = (1/4) - (1/36)

    = 0.2222

    - RH = 1.0968 * 107 m-1, to find a value for 1/λ. The formula and the example calculation gives:

    1/λ = RH [ (1/2²) - (1 / n²₂) ]

    = 1.0968 * 107 m-1 * 3/16

    = 2437090 m-1

    - λ = 1 / 2437090 m-1

    = 4.10 * 10-7 m

    = 410 nanometers

    n=5 to n=2

    - The principal quantum numbers are 2 and 5.

    - (1/2²) - (1 / n²₂)

    For n₂ = 5, you get:

    (1/2²) - (1 / n²₂) = (1/2²) - (1 / 5²)

    = (1/4) - (1/25)

    = 0.21

    - RH = 1.0968 * 107 m-1, to find a value for 1/λ. The formula and the example calculation gives:

    1/λ = RH [ (1/2²) - (1 / n²₂) ]

    = 1.0968 * 107 m-1 * 0.21

    = 2303280 m-1

    - λ = 1 / 2303280 m-1

    = 4.34 * 10-7 m

    = 434 nanometers

    n=4 to n=2

    - The principal quantum numbers are 2 and 4.

    - (1/2²) - (1 / n²₂)

    For n₂ = 4, you get:

    (1/2²) - (1 / n²₂) = (1/2²) - (1 / 4²)

    = (1/4) - (1/16)

    = 0.1875

    - RH = 1.0968 * 107 m-1, to find a value for 1/λ. The formula and the example calculation gives:

    1/λ = RH [ (1/2²) - (1 / n²₂) ]

    = 1.0968 * 107 m-1 * 0.1875

    = 2056500 m-1

    - λ = 1 / 2056500 m-1

    = 4.86 * 10-7 m

    = 486 nanometers

    n=3 to n=2

    - The principal quantum numbers are 2 and 3.

    - (1/2²) - (1 / n²₂)

    For n₂ = 3, you get:

    (1/2²) - (1 / n²₂) = (1/2²) - (1 / 3²)

    = (1/4) - (1/9)

    = 0.13889

    - RH = 1.0968 * 107 m-1, to find a value for 1/λ. The formula and the example calculation gives:

    1/λ = RH [ (1/2²) - (1 / n²₂) ]

    = 1.0968 * 107 m-1 * 0.13889

    = 1523345 m-1

    - λ = 1 / 1523345 m-1

    = 6.56 * 10-7 m

    = 656 nanometers
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