Question

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.
2. circle the correct bold word for each choice

in the experiment, the metal cations’ electrons were initially in the (ground, excited) state. when placed in the flame, the electrons then (absorbed, emitted) energy in the form of (electricity, heat). when this occurred, electrons made transitions from (low, high) energy levels to (low, high) energy levels. the electrons were then in the (ground, excited) state. the electrons in these metals then made transitions from (low, high) energy levels to (low, high) energy levels, resulting in an (absorption, emission) of energy as (electricity, heat, light)

Explanation:

Step1: Recall the energy formula

The formula for the energy of a photon is \( E = \frac{hc}{\lambda} \), where \( h = 6.626\times10^{-34}\, \text{J·s} \) (Planck's constant), \( c = 3.0\times10^{8}\, \text{m/s} \) (speed of light), and \( \lambda \) is the wavelength in meters.

Step2: Choose an ion (e.g., \( \text{Na}^+ \))

For \( \text{Na}^+ \), the wavelength \( \lambda = 6.30\times10^{-7}\, \text{m} \).

Step3: Substitute values into the formula

\( E = \frac{(6.626\times10^{-34}\, \text{J·s})(3.0\times10^{8}\, \text{m/s})}{6.30\times10^{-7}\, \text{m}} \)

First, calculate the numerator: \( (6.626\times10^{-34})(3.0\times10^{8}) = 1.9878\times10^{-25}\, \text{J·m} \)

Then divide by the wavelength: \( \frac{1.9878\times10^{-25}}{6.30\times10^{-7}} \approx 3.155\times10^{-19}\, \text{J} \)

Answer:

For \( \text{Na}^+ \), the energy is approximately \( 3.16\times10^{-19}\, \text{J} \) (values may vary slightly with more precise calculations).