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 | 3.16×10⁻¹⁹
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}{multirow{2}{*}{calcium}}
unknown c | red, orange | multicolumn{3}{c}{}

Explanation:

Step1: Recall Energy-Wavelength Formula

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

Step2: Calculate Energy for Potassium ($K^+$)

Given $\lambda = 640\ \text{nm} = 640\times10^{-9}\ \text{m}$.
Substitute into the formula:
$E = \frac{(6.626\times10^{-34})(3.0\times10^{8})}{640\times10^{-9}}$
First, multiply numerator: $(6.626\times10^{-34})(3.0\times10^{8}) = 1.9878\times10^{-25}$
Then divide by denominator: $\frac{1.9878\times10^{-25}}{6.4\times10^{-7}} \approx 3.106\times10^{-19}\ \text{J}$

Step3: Calculate Energy for Lithium ($Li^+$)

$\lambda = 670\ \text{nm} = 670\times10^{-9}\ \text{m}$.
$E = \frac{(6.626\times10^{-34})(3.0\times10^{8})}{670\times10^{-9}}$
Numerator: $1.9878\times10^{-25}$
Denominator: $6.7\times10^{-7}$
$E \approx \frac{1.9878\times10^{-25}}{6.7\times10^{-7}} \approx 2.967\times10^{-19}\ \text{J}$

Step4: Calculate Energy for Calcium ($Ca^{2+}$)

$\lambda = 650\ \text{nm} = 650\times10^{-9}\ \text{m}$.
$E = \frac{(6.626\times10^{-34})(3.0\times10^{8})}{650\times10^{-9}}$
Numerator: $1.9878\times10^{-25}$
Denominator: $6.5\times10^{-7}$
$E \approx \frac{1.9878\times10^{-25}}{6.5\times10^{-7}} \approx 3.058\times10^{-19}\ \text{J}$

Step5: Calculate Energy for Strontium ($Sr^{2+}$)

$\lambda = 670\ \text{nm}$ (same as Lithium, so energy is also $\approx 2.967\times10^{-19}\ \text{J}$)

Step6: Calculate Energy for Barium ($Ba^{2+}$)

$\lambda = 610\ \text{nm} = 610\times10^{-9}\ \text{m}$.
$E = \frac{(6.626\times10^{-34})(3.0\times10^{8})}{610\times10^{-9}}$
Numerator: $1.9878\times10^{-25}$
Denominator: $6.1\times10^{-7}$
$E \approx \frac{1.9878\times10^{-25}}{6.1\times10^{-7}} \approx 3.259\times10^{-19}\ \text{J}$

Step7: Calculate Energy for Copper ($Cu^{2+}$)

$\lambda = 550\ \text{nm} = 550\times10^{-9}\ \text{m}$.
$E = \frac{(6.626\times10^{-34})(3.0\times10^{8})}{550\times10^{-9}}$
Numerator: $1.9878\times10^{-25}$
Denominator: $5.5\times10^{-7}$
$E \approx \frac{1.9878\times10^{-25}}{5.5\times10^{-7}} \approx 3.614\times10^{-19}\ \text{J}$

Answer:

• Potassium ($K^+$): $\approx 3.11\times10^{-19}\ \text{J}$
• Lithium ($Li^+$): $\approx 2.97\times10^{-19}\ \text{J}$
• Calcium ($Ca^{2+}$): $\approx 3.06\times10^{-19}\ \text{J}$
• Strontium ($Sr^{2+}$): $\approx 2.97\times10^{-19}\ \text{J}$
• Barium ($Ba^{2+}$): $\approx 3.26\times10^{-19}\ \text{J}$
• Copper ($Cu^{2+}$): $\approx 3.61\times10^{-19}\ \text{J}$