To solve the problem where an electron transitions from the [tex]\( n=4 \)[/tex] energy level to the [tex]\( n=3 \)[/tex] energy level in a hydrogen atom, we need to calculate the energy difference ([tex]\(\Delta E\)[/tex]) and the corresponding frequency ([tex]\( \nu \)[/tex]) of the emitted photon.
### Step 1: Calculate the Energy Difference ([tex]\(\Delta E\)[/tex])
We use the formula:
[tex]\[
\Delta E = -2.18 \times 10^{-18} \, \text{J} \left( \frac{1}{n_\text{final}^2} - \frac{1}{n_\text{initial}^2} \right)
\][/tex]
Let's plug in the values:
- [tex]\( n_\text{initial} = 4 \)[/tex]
- [tex]\( n_\text{final} = 3 \)[/tex]
So, the calculation becomes:
[tex]\[
\Delta E = -2.18 \times 10^{-18} \, \text{J} \left( \frac{1}{3^2} - \frac{1}{4^2} \right)
\][/tex]
[tex]\[
\Delta E = -2.18 \times 10^{-18} \, \text{J} \left( \frac{1}{9} - \frac{1}{16} \right)
\][/tex]
Simplify the fractions:
[tex]\[
\frac{1}{9} = 0.1111\ldots \quad \text{and} \quad \frac{1}{16} = 0.0625
\][/tex]
Then:
[tex]\[
\Delta E = -2.18 \times 10^{-18} \, \text{J} \left( 0.1111\ldots - 0.0625 \right)
\][/tex]
[tex]\[
\Delta E = -2.18 \times 10^{-18} \, \text{J} \left( 0.0486\ldots \right)
\][/tex]
The energy difference is:
[tex]\[
\Delta E = -1.05972222222 \times 10^{-19} \, \text{J}
\][/tex]
### Step 2: Calculate the Frequency ([tex]\( \nu \)[/tex])
Using the formula:
[tex]\[
\Delta E = h \cdot \nu
\][/tex]
Where [tex]\( h \)[/tex] is the Planck's constant ([tex]\( h = 6.62607015 \times 10^{-34} \, \text{J} \cdot \text{s} \)[/tex]). To find the frequency ([tex]\( \nu \)[/tex]):
[tex]\[
\nu = \frac{\Delta E}{h}
\][/tex]
Substitute the known values:
[tex]\[
\nu = \frac{1.05972222222 \times 10^{-19} \, \text{J}}{6.62607015 \times 10^{-34} \, \text{J} \cdot \text{s}}
\][/tex]
Perform the division:
[tex]\[
\nu = 159932237092633.56 \, \text{Hz}
\][/tex]
### Step 3: Express the Frequency to Three Significant Figures
Round the result to three significant figures:
[tex]\[
\nu \approx 1.60 \times 10^{14} \, \text{Hz}
\][/tex]
### Solution Summary
- The energy difference ([tex]\(\Delta E\)[/tex]) is [tex]\(-1.06 \times 10^{-19} \, \text{J}\)[/tex] (approximated to three significant figures).
- The frequency ([tex]\(\nu\)[/tex]) of the emitted photon is [tex]\( 1.60 \times 10^{14} \, \text{Hz} \)[/tex] (to three significant figures).
