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

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