5. below is data collected on the amount of energy emitted by the electrons of hydrogen, oxygen, and lithium. convert their energies to wavelengths in nanometers, then using the table of wavelengths and corresponding colors, draw the emission spectra of these elements using color pencils. you must show all work using dimensional analysis. include units in your work and answers. round your final answer to the correct number of significant figures.
hydrogen:
show your work here:

Question

Accepted Answer

# Explanation:
## Step1: Recall the formula
The formula relating energy $E$ and wavelength $\lambda$ is $E = \frac{hc}{\lambda}$, where $h = 6.626\times10^{-34}\text{ J}\cdot\text{s}$ (Planck's constant) and $c= 3\times 10^{8}\text{ m/s}$ (speed of light). Rearranging for $\lambda$, we get $\lambda=\frac{hc}{E}$. Also, $1\text{ m}=10^{9}\text{ nm}$.

## Step2: Calculate for Hydrogen's first energy value ($E = 3.03\times 10^{-19}\text{ J}$)
$$
\begin{align*}
\lambda&=\frac{6.626\times 10^{-34}\text{ J}\cdot\text{s}\times3\times 10^{8}\text{ m/s}}{3.03\times 10^{-19}\text{ J}}\times10^{9}\text{ nm/m}\
&=\frac{19.878\times 10^{-26}}{3.03\times 10^{-19}}\times10^{9}\text{ nm}\
&=\frac{19.878}{3.03}\times10^{-26 + 19+9}\text{ nm}\
&=6.56\times10^{2}\text{ nm}
\end{align*}
$$

## Step3: Calculate for Hydrogen's second energy value ($E = 4.09\times 10^{-19}\text{ J}$)
$$
\begin{align*}
\lambda&=\frac{6.626\times 10^{-34}\text{ J}\cdot\text{s}\times3\times 10^{8}\text{ m/s}}{4.09\times 10^{-19}\text{ J}}\times10^{9}\text{ nm/m}\
&=\frac{19.878\times 10^{-26}}{4.09\times 10^{-19}}\times10^{9}\text{ nm}\
&=\frac{19.878}{4.09}\times10^{-26 + 19+9}\text{ nm}\
&=4.86\times10^{2}\text{ nm}
\end{align*}
$$

## Step4: Calculate for Hydrogen's third energy value ($E = 4.58\times 10^{-19}\text{ J}$)
$$
\begin{align*}
\lambda&=\frac{6.626\times 10^{-34}\text{ J}\cdot\text{s}\times3\times 10^{8}\text{ m/s}}{4.58\times 10^{-19}\text{ J}}\times10^{9}\text{ nm/m}\
&=\frac{19.878\times 10^{-26}}{4.58\times 10^{-19}}\times10^{9}\text{ nm}\
&=\frac{19.878}{4.58}\times10^{-26 + 19+9}\text{ nm}\
&=4.34\times10^{2}\text{ nm}
\end{align*}
$$

## Step5: Calculate for Hydrogen's fourth energy value ($E = 4.85\times 10^{-19}\text{ J}$)
$$
\begin{align*}
\lambda&=\frac{6.626\times 10^{-34}\text{ J}\cdot\text{s}\times3\times 10^{8}\text{ m/s}}{4.85\times 10^{-19}\text{ J}}\times10^{9}\text{ nm/m}\
&=\frac{19.878\times 10^{-26}}{4.85\times 10^{-19}}\times10^{9}\text{ nm}\
&=\frac{19.878}{4.85}\times10^{-26 + 19+9}\text{ nm}\
&=4.10\times10^{2}\text{ nm}
\end{align*}
$$

# Answer:
For Hydrogen, the wavelengths (in nm) are approximately $656$, $486$, $434$, $410$ (corresponding to colors in the visible spectrum as per the given wavelength - color table).

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QUESTION IMAGE

Question

Explanation:

Step1: Recall the formula

Step2: Calculate for Hydrogen's first energy value (\(E = 3.03\times 10^{-19}\text{ J}\))

Step3: Calculate for Hydrogen's second energy value (\(E = 4.09\times 10^{-19}\text{ J}\))

Step4: Calculate for Hydrogen's third energy value (\(E = 4.58\times 10^{-19}\text{ J}\))

Step5: Calculate for Hydrogen's fourth energy value (\(E = 4.85\times 10^{-19}\text{ J}\))

Answer: