Question

solve this problem including all necessary steps
An excited H atom emits light with a frequency of 1.141x1014s
-1
to reach the n=4 energy
level. In what principle quantum number did the electron originally reside?

Answer 1

To find the original principal quantum number of the electron in a hydrogen atom that emits light with a frequency of \(1.141 \times 10^{14} \, \text{s}^{-1}\) as it falls to the \(n=4\) energy level, we will use the Rydberg formula for hydrogen:

\[
\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)
\]

where:
- \(R_H\) is the Rydberg constant (\(1.097 \times 10^7 \, \text{m}^{-1}\)),
- \(n_1\) is the principal quantum number of the lower energy level,
- \(n_2\) is the principal quantum number of the higher energy level,
- \(\lambda\) is the wavelength of emitted light.

### Step 1: Calculate the wavelength (\(\lambda\))

Since we have the frequency (\(f\)), we can find the wavelength using the relationship:

\[
\lambda = \frac{c}{f}
\]

where \(c\) is the speed of light (\(3.00 \times 10^8 \, \text{m/s}\)).

Substituting the values into the equation:

\[
\lambda = \frac{3.00 \times 10^8 \, \text{m/s}}{1.141 \times 10^{14} \, \text{s}^{-1}}
\]

Calculating this gives:

\[
\lambda \approx 2.63 \times 10^{-6} \, \text{m} = 2630 \, \text{nm}
\]

### Step 2: Use the Rydberg formula

Now that we have \(\lambda\), we can substitute it into the Rydberg equation. First, we need to convert \(\lambda\) into meters in the form of \(\frac{1}{\lambda}\):

\[
\frac{1}{\lambda} = \frac{1}{2.63 \times 10^{-6}} \approx 3.80 \times 10^5 \, \text{m}^{-1}
\]

Now we will plug it into the Rydberg formula:

\[
3.80 \times 10^5 = R_H \left( \frac{1}{n_1^2} - \frac{1}{4^2} \right)
\]

Substituting \(R_H\):

\[
3.80 \times 10^5 = 1.097 \times 10^7 \left( \frac{1}{n_1^2} - \frac{1}{16} \right)
\]

### Step 3: Solve for \(\frac{1}{n_1^2}\)

We can simplify this equation:

\[
\frac{1}{n_1^2} - \frac{1}{16} = \frac{3.80 \times 10^5}{1.097 \times 10^7}
\]

Calculating the right side:

\[
\frac{3.80 \times 10^5}{1.097 \times 10^7} \approx 0.0346
\]

So now we have:

\[
\frac{1}{n_1^2} - \frac{1}{16} = 0.0346
\]

### Step 4: Isolate \(\frac{1}{n_1^2}\)

Adding \(\frac{1}{16}\) to both sides:

\[
\frac{1}{n_1^2} = 0.0346 + \frac{1}{16}
\]

Calculating \(\frac{1}{16}\):

\[
\frac{1}{16} = 0.0625
\]

So,

\[
\frac{1}{n_1^2} = 0.0346 + 0.0625 \approx 0.0971
\]

### Step 5: Find \(n_1\)

Now we take the reciprocal and then the square root:

\[
n_1^2 = \frac{1}{0.0971} \approx 10.31 \implies n_1 \approx \sqrt{10.31} \approx 3.21
\]

Since \(n_1\) must be a whole number, we round it to the nearest whole number, which is:

\[
\boxed{3}
\]

Thus, the electron originally resided in the \(n = 3\) energy level.