Which electronic transition in a hydrogen atom, starting from the orbit n=7, will produce infrared light of wavelength 2170 nm?
`(Given: R_(H)=1.09677xx10^(7)M^(-1))`

Options

A. `n=7 " to " n=4`
B. from `n=2` to `n=4`
C. from `n=5` to `n=1`
D. from `n=3` to `n=5`

Answer

To determine which electronic transition in a hydrogen atom starting from the orbit \( n = 7 \) will produce infrared light of wavelength \( 2170 \, \text{nm} \), we can use the Rydberg formula for hydrogen:

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

where:
- \( \lambda \) is the wavelength of the emitted light,
- \( R_H \) is the Rydberg constant for hydrogen, \( R_H = 1.09677 \times 10^7 \, \text{m}^{-1} \),
- \( n_1 \) is the lower energy level,
- \( n_2 \) is the higher energy level (in this case, \( n_2 = 7 \)).

### Step-by-step Solution:

1. **Convert Wavelength to Meters**:
The given wavelength is \( 2170 \, \text{nm} \). We need to convert this to meters:
\[ \lambda = 2170 \, \text{nm} = 2170 \times 10^{-9} \, \text{m} \]

2. **Substitute Values into the Rydberg Formula**:
We know \( n_2 = 7 \) and \( R_H = 1.09677 \times 10^7 \, \text{m}^{-1} \). Substitute these values into the Rydberg formula:
\[ \frac{1}{2170 \times 10^{-9}} = 1.09677 \times 10^7 \left( \frac{1}{n_1^2} - \frac{1}{7^2} \right) \]

3. **Calculate \( \frac{1}{\lambda} \)**:
Calculate \( \frac{1}{2170 \times 10^{-9}} \):
\[ \frac{1}{2170 \times 10^{-9}} \approx 4.608 \times 10^6 \, \text{m}^{-1} \]

4. **Set Up the Equation**:
Now we can set up the equation:
\[ 4.608 \times 10^6 = 1.09677 \times 10^7 \left( \frac{1}{n_1^2} - \frac{1}{49} \right) \]

5. **Isolate \( \frac{1}{n_1^2} \)**:
Rearranging gives:
\[ \frac{1}{n_1^2} - \frac{1}{49} = \frac{4.608 \times 10^6}{1.09677 \times 10^7} \]
Calculate the right side:
\[ \frac{4.608 \times 10^6}{1.09677 \times 10^7} \approx 0.420 \]
Therefore:
\[ \frac{1}{n_1^2} = 0.420 + \frac{1}{49} \]

6. **Calculate \( \frac{1}{49} \)**:
\[ \frac{1}{49} \approx 0.0204 \]
So:
\[ \frac{1}{n_1^2} = 0.420 + 0.0204 \approx 0.4404 \]

7. **Find \( n_1^2 \)**:
Taking the reciprocal:
\[ n_1^2 \approx \frac{1}{0.4404} \approx 2.27 \]

8. **Calculate \( n_1 \)**:
Taking the square root:
\[ n_1 \approx \sqrt{2.27} \approx 1.5 \]
Since \( n_1 \) must be an integer, we round it to \( n_1 = 4 \).

9. **Conclusion**:
The electronic transition occurs from \( n_2 = 7 \) to \( n_1 = 4 \).

### Final Answer:
The electronic transition in a hydrogen atom that produces infrared light of wavelength \( 2170 \, \text{nm} \) is from \( n = 7 \) to \( n = 4 \).