Answer :
To determine the excited state [tex]\( n_2 \)[/tex] for a hydrogen atom when a wavelength of 97.3 nm is emitted with [tex]\( n_1 = 1 \)[/tex], we can use the Rydberg formula for hydrogen emission spectra:
[tex]\[
\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)
\][/tex]
Here, [tex]\( \lambda \)[/tex] is the wavelength of the emitted light, [tex]\( R \)[/tex] is the Rydberg constant, and [tex]\( n_1 \)[/tex] and [tex]\( n_2 \)[/tex] are the principal quantum numbers of the initial and final energy levels respectively. For hydrogen, the Rydberg constant [tex]\( R \approx 1.097 \times 10^7 \, \text{m}^{-1} \)[/tex].
Step-by-Step Solution:
1. Convert Wavelength to Meters:
- Given [tex]\(\lambda = 97.3 \, \text{nm}\)[/tex], convert this to meters:
[tex]\[
\lambda = 97.3 \times 10^{-9} \, \text{m}
\][/tex]
2. Apply the Rydberg Formula:
- The formula is:
[tex]\[
\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)
\][/tex]
- Substitute the known values:
[tex]\[
\frac{1}{97.3 \times 10^{-9}} = 1.097 \times 10^7 \left( \frac{1}{1^2} - \frac{1}{n_2^2} \right)
\][/tex]
3. Solve for [tex]\( \frac{1}{n_2^2} \)[/tex]:
- Calculate the left side:
[tex]\[
\frac{1}{97.3 \times 10^{-9}} \approx 1.028 \times 10^{7} \, \text{m}^{-1}
\][/tex]
- Rearrange the formula:
[tex]\[
1.028 \times 10^{7} = 1.097 \times 10^7 \left( 1 - \frac{1}{n_2^2} \right)
\][/tex]
4. Calculate [tex]\( n_2^2: \)[/tex]
- Divide both sides by [tex]\( 1.097 \times 10^7 \)[/tex]:
[tex]\[
\frac{1.028 \times 10^{7}}{1.097 \times 10^7} = 1 - \frac{1}{n_2^2}
\][/tex]
- Calculate the fraction:
[tex]\[
\approx 0.9374 = 1 - \frac{1}{n_2^2}
\][/tex]
- Therefore:
[tex]\[
\frac{1}{n_2^2} = 1 - 0.9374 = 0.0626
\][/tex]
- Take the reciprocal:
[tex]\[
n_2^2 \approx \frac{1}{0.0626} \approx 15.97
\][/tex]
- Look for the integer [tex]\( n_2 \)[/tex]:
[tex]\[
n_2 \approx 4
\][/tex]
5. Select from Given Options:
- The closest integer to our calculation is 4.
- Therefore, the excited state [tex]\( n_2 \)[/tex] is 4.
The correct option is D. 4.
[tex]\[
\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)
\][/tex]
Here, [tex]\( \lambda \)[/tex] is the wavelength of the emitted light, [tex]\( R \)[/tex] is the Rydberg constant, and [tex]\( n_1 \)[/tex] and [tex]\( n_2 \)[/tex] are the principal quantum numbers of the initial and final energy levels respectively. For hydrogen, the Rydberg constant [tex]\( R \approx 1.097 \times 10^7 \, \text{m}^{-1} \)[/tex].
Step-by-Step Solution:
1. Convert Wavelength to Meters:
- Given [tex]\(\lambda = 97.3 \, \text{nm}\)[/tex], convert this to meters:
[tex]\[
\lambda = 97.3 \times 10^{-9} \, \text{m}
\][/tex]
2. Apply the Rydberg Formula:
- The formula is:
[tex]\[
\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)
\][/tex]
- Substitute the known values:
[tex]\[
\frac{1}{97.3 \times 10^{-9}} = 1.097 \times 10^7 \left( \frac{1}{1^2} - \frac{1}{n_2^2} \right)
\][/tex]
3. Solve for [tex]\( \frac{1}{n_2^2} \)[/tex]:
- Calculate the left side:
[tex]\[
\frac{1}{97.3 \times 10^{-9}} \approx 1.028 \times 10^{7} \, \text{m}^{-1}
\][/tex]
- Rearrange the formula:
[tex]\[
1.028 \times 10^{7} = 1.097 \times 10^7 \left( 1 - \frac{1}{n_2^2} \right)
\][/tex]
4. Calculate [tex]\( n_2^2: \)[/tex]
- Divide both sides by [tex]\( 1.097 \times 10^7 \)[/tex]:
[tex]\[
\frac{1.028 \times 10^{7}}{1.097 \times 10^7} = 1 - \frac{1}{n_2^2}
\][/tex]
- Calculate the fraction:
[tex]\[
\approx 0.9374 = 1 - \frac{1}{n_2^2}
\][/tex]
- Therefore:
[tex]\[
\frac{1}{n_2^2} = 1 - 0.9374 = 0.0626
\][/tex]
- Take the reciprocal:
[tex]\[
n_2^2 \approx \frac{1}{0.0626} \approx 15.97
\][/tex]
- Look for the integer [tex]\( n_2 \)[/tex]:
[tex]\[
n_2 \approx 4
\][/tex]
5. Select from Given Options:
- The closest integer to our calculation is 4.
- Therefore, the excited state [tex]\( n_2 \)[/tex] is 4.
The correct option is D. 4.
