Explanation:
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
is a Lewis acid due to the unfilled valence shell on carbon atom (possessing positive charge) it accepts electron readily.
is not a Lewis acid due to the presence of negative charge which represents that there are electrons available for donation.
is not a Lewis acid due to the presence of lone pair on nitrogen which is easily donated to the acceptors.
is a Lewis acid due to the availability of vacant p-orbital on boron atom.
