Question

fill in the color column while your bunsen burner is on. when your burner is off, complete the rest of the table
metallic ion | color of flame (be specific: orange, yellow - orange, red - orange) | approximate wavelength (nm) | wavelength (m) | energy (j)
na⁺, sodium | orange | 630 | 6.30×10⁻⁷ |
k⁺, potassium | red, orange | 640 | 6.40×10⁻⁷ |
li⁺, lithium | red | 670 | 6.70×10⁻⁷ |
ca²⁺, calcium | red, orange | 650 | 6.50×10⁻⁷ |
sr²⁺, strontium | red | 670 | 6.70×10⁻⁷ |
ba²⁺, barium | yellow | 610 | 6.10×10⁻⁷ |
cu²⁺, copper | green | 550 | 5.50×10⁻⁷ |
unknown a | | multicolumn{3}{c}{list just the name of each unknown ion here!}
unknown b | | multicolumn{3}{c}{
unknown c | red, orange | multicolumn{3}{c}{calcium}
analysis and conclusions

1. finish filling out the shaded boxes in the data table. show your work below for one ion.

Explanation:

Step1: Recall Energy Formula

The formula for the energy of a photon is \( E = \frac{hc}{\lambda} \), where \( h = 6.626 \times 10^{-34} \, \text{J·s} \) (Planck's constant), \( c = 3.0 \times 10^{8} \, \text{m/s} \) (speed of light), and \( \lambda \) is the wavelength in meters.

Step2: Choose an Ion (e.g., Sodium, \( \lambda = 630 \, \text{nm} = 630 \times 10^{-9} \, \text{m} \))

Convert wavelength to meters: \( \lambda = 630 \times 10^{-9} \, \text{m} = 6.3 \times 10^{-7} \, \text{m} \).

Step3: Calculate Energy

Substitute into the formula: \( E = \frac{(6.626 \times 10^{-34})(3.0 \times 10^{8})}{6.3 \times 10^{-7}} \).
First, multiply numerator: \( (6.626 \times 10^{-34})(3.0 \times 10^{8}) = 1.9878 \times 10^{-25} \).
Then divide by \( 6.3 \times 10^{-7} \): \( E = \frac{1.9878 \times 10^{-25}}{6.3 \times 10^{-7}} \approx 3.155 \times 10^{-19} \, \text{J} \approx 3.16 \times 10^{-19} \, \text{J} \) (matches the given value for Na⁺, verifying the method).
For other ions, follow the same steps: convert \( \lambda \) (nm to m: \( \lambda (\text{m}) = \lambda (\text{nm}) \times 10^{-9} \)), then use \( E = \frac{hc}{\lambda} \).

Answer:

For example, for Sodium (\( \text{Na}^+ \)):
Energy calculation: \( E = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34} \, \text{J·s})(3.0 \times 10^{8} \, \text{m/s})}{630 \times 10^{-9} \, \text{m}} \approx 3.16 \times 10^{-19} \, \text{J} \) (consistent with the table).
For Potassium (\( \text{K}^+ \), \( \lambda = 640 \, \text{nm} = 6.4 \times 10^{-7} \, \text{m} \)):
\( E = \frac{(6.626 \times 10^{-34})(3.0 \times 10^{8})}{6.4 \times 10^{-7}} \approx \frac{1.9878 \times 10^{-25}}{6.4 \times 10^{-7}} \approx 3.106 \times 10^{-19} \, \text{J} \approx 3.11 \times 10^{-19} \, \text{J} \).
(Repeat for each ion using their respective wavelengths to fill the "Energy (J)" column.)