Wavelength, Frequency & Energy Worksheet with detailed calculations and answers.

Problem Analysis:

The worksheet involves calculations related to the relationships between wavelength (\(\lambda\)), frequency (\(f\)), energy (\(E\)), and the speed of light (\(c\)). The key equations used are:

1. Wavelength and Frequency Relationship:
\[
c = \lambda f
\]
where \(c = 3.00 \times 10^8 \, \text{m/s}\).

2. Energy and Frequency Relationship (Planck's Equation):
\[
E = h f
\]
where \(h = 6.626 \times 10^{-34} \, \text{J·s}\).

3. Energy and Wavelength Relationship:
\[
E = \frac{hc}{\lambda}
\]

Solution to Each Problem:

#### Problem 1:
What is the frequency of a spectral line having the wavelength \(4.8 \times 10^{-7} \, \text{m}\)?

Using \(c = \lambda f\):
\[
f = \frac{c}{\lambda} = \frac{3.00 \times 10^8 \, \text{m/s}}{4.8 \times 10^{-7} \, \text{m}}
\]
\[
f = \frac{3.00 \times 10^8}{4.8 \times 10^{-7}} = \frac{3.00}{4.8} \times 10^{8 + 7} = 0.625 \times 10^{15} \, \text{Hz}
\]
\[
f = 6.25 \times 10^{14} \, \text{Hz}
\]

#### Problem 2:
What is the frequency of a spectral line having the wavelength \(580 \, \text{nm}\)?

Convert \(580 \, \text{nm}\) to meters:
\[
580 \, \text{nm} = 580 \times 10^{-9} \, \text{m} = 5.80 \times 10^{-7} \, \text{m}
\]
Using \(c = \lambda f\):
\[
f = \frac{c}{\lambda} = \frac{3.00 \times 10^8 \, \text{m/s}}{5.80 \times 10^{-7} \, \text{m}}
\]
\[
f = \frac{3.00}{5.80} \times 10^{8 + 7} = 0.5172 \times 10^{15} \, \text{Hz}
\]
\[
f \approx 5.17 \times 10^{14} \, \text{Hz}
\]

#### Problem 3:
What is the wavelength of a spectral line having the frequency \(6.2 \times 10^{14} \, \text{Hz}\)?

Using \(c = \lambda f\):
\[
\lambda = \frac{c}{f} = \frac{3.00 \times 10^8 \, \text{m/s}}{6.2 \times 10^{14} \, \text{Hz}}
\]
\[
\lambda = \frac{3.00}{6.2} \times 10^{8 - 14} = 0.4839 \times 10^{-6} \, \text{m}
\]
\[
\lambda \approx 4.84 \times 10^{-7} \, \text{m}
\]

#### Problem 4:
What is the energy jump indicated by a spectral line with frequency \(6.2 \times 10^{14} \, \text{Hz}\)?

Using \(E = h f\):
\[
E = (6.626 \times 10^{-34} \, \text{J·s}) \times (6.2 \times 10^{14} \, \text{Hz})
\]
\[
E = 6.626 \times 6.2 \times 10^{-34 + 14} = 41.0812 \times 10^{-20} \, \text{J}
\]
\[
E \approx 4.11 \times 10^{-19} \, \text{J}
\]

#### Problem 5:
What is the frequency of a spectral line with energy \(5.9 \times 10^{-19} \, \text{J}\)?

Using \(E = h f\):
\[
f = \frac{E}{h} = \frac{5.9 \times 10^{-19} \, \text{J}}{6.626 \times 10^{-34} \, \text{J·s}}
\]
\[
f = \frac{5.9}{6.626} \times 10^{-19 + 34} = 0.8907 \times 10^{15} \, \text{Hz}
\]
\[
f \approx 8.91 \times 10^{14} \, \text{Hz}
\]

#### Problem 6:
What is the energy of a spectral line at \(6.8 \times 10^{-7} \, \text{m}\)?

Using \(E = \frac{hc}{\lambda}\):
\[
E = \frac{(6.626 \times 10^{-34} \, \text{J·s})(3.00 \times 10^8 \, \text{m/s})}{6.8 \times 10^{-7} \, \text{m}}
\]
\[
E = \frac{6.626 \times 3.00}{6.8} \times 10^{-34 + 8 + 7} = \frac{19.878}{6.8} \times 10^{-19}
\]
\[
E \approx 2.92 \times 10^{-19} \, \text{J}
\]

#### Problem 7:
What is the energy of a spectral line at \(340 \, \text{nm}\)?

Convert \(340 \, \text{nm}\) to meters:
\[
340 \, \text{nm} = 340 \times 10^{-9} \, \text{m} = 3.40 \times 10^{-7} \, \text{m}
\]
Using \(E = \frac{hc}{\lambda}\):
\[
E = \frac{(6.626 \times 10^{-34} \, \text{J·s})(3.00 \times 10^8 \, \text{m/s})}{3.40 \times 10^{-7} \, \text{m}}
\]
\[
E = \frac{6.626 \times 3.00}{3.40} \times 10^{-34 + 8 + 7} = \frac{19.878}{3.40} \times 10^{-19}
\]
\[
E \approx 5.85 \times 10^{-19} \, \text{J}
\]

#### Problem 8:
What is the wavelength of a spectral line having energy \(4.4 \times 10^{-19} \, \text{J}\)?

Using \(E = \frac{hc}{\lambda}\):
\[
\lambda = \frac{hc}{E} = \frac{(6.626 \times 10^{-34} \, \text{J·s})(3.00 \times 10^8 \, \text{m/s})}{4.4 \times 10^{-19} \, \text{J}}
\]
\[
\lambda = \frac{6.626 \times 3.00}{4.4} \times 10^{-34 + 8 + 19} = \frac{19.878}{4.4} \times 10^{-7}
\]
\[
\lambda \approx 4.52 \times 10^{-7} \, \text{m} = 452 \, \text{nm}
\]

#### Problem 9:
What is the wavelength of a spectral line having a frequency of \(8.7 \times 10^{14} \, \text{Hz}\)?

Using \(c = \lambda f\):
\[
\lambda = \frac{c}{f} = \frac{3.00 \times 10^8 \, \text{m/s}}{8.7 \times 10^{14} \, \text{Hz}}
\]
\[
\lambda = \frac{3.00}{8.7} \times 10^{8 - 14} = 0.3448 \times 10^{-6} \, \text{m}
\]
\[
\lambda \approx 3.45 \times 10^{-7} \, \text{m} = 345 \, \text{nm}
\]

#### Problem 10:
What is the energy of a spectral line having a frequency of \(7.2 \times 10^{14} \, \text{Hz}\)?

Using \(E = h f\):
\[
E = (6.626 \times 10^{-34} \, \text{J·s}) \times (7.2 \times 10^{14} \, \text{Hz})
\]
\[
E = 6.626 \times 7.2 \times 10^{-34 + 14} = 47.7192 \times 10^{-20} \, \text{J}
\]
\[
E \approx 4.77 \times 10^{-19} \, \text{J}
\]

#### Problem 11:
What is the frequency of a spectral line having an energy of \(5.5 \times 10^{-19} \, \text{J}\)?

Using \(E = h f\):
\[
f = \frac{E}{h} = \frac{5.5 \times 10^{-19} \, \text{J}}{6.626 \times 10^{-34} \, \text{J·s}}
\]
\[
f = \frac{5.5}{6.626} \times 10^{-19 + 34} = 0.8299 \times 10^{15} \, \text{Hz}
\]
\[
f \approx 8.30 \times 10^{14} \, \text{Hz}
\]

#### Problem 12:
What is the wavelength of a spectral line having the energy of \(3.2 \times 10^{-19} \, \text{J}\)?

Using \(E = \frac{hc}{\lambda}\):
\[
\lambda = \frac{hc}{E} = \frac{(6.626 \times 10^{-34} \, \text{J·s})(3.00 \times 10^8 \, \text{m/s})}{3.2 \times 10^{-19} \, \text{J}}
\]
\[
\lambda = \frac{6.626 \times 3.00}{3.2} \times 10^{-34 + 8 + 19} = \frac{19.878}{3.2} \times 10^{-7}
\]
\[
\lambda \approx 6.21 \times 10^{-7} \, \text{m} = 621 \, \text{nm}
\]

Final Answers:

\[
\boxed{
\begin{aligned}
1. & \, 6.25 \times 10^{14} \, \text{Hz} \\
2. & \, 5.17 \times 10^{14} \, \text{Hz} \\
3. & \, 4.84 \times 10^{-7} \, \text{m} \\
4. & \, 4.11 \times 10^{-19} \, \text{J} \\
5. & \, 8.91 \times 10^{14} \, \text{Hz} \\
6. & \, 2.92 \times 10^{-19} \, \text{J} \\
7. & \, 5.85 \times 10^{-19} \, \text{J} \\
8. & \, 452 \, \text{nm} \\
9. & \, 345 \, \text{nm} \\
10. & \, 4.77 \times 10^{-19} \, \text{J} \\
11. & \, 8.30 \times 10^{14} \, \text{Hz} \\
12. & \, 621 \, \text{nm}
\end{aligned}
}
\]