To determine the wavelength of light emitted when an electron in a hydrogen atom transitions from the energy level [tex]\(n=5\)[/tex] to [tex]\(n=1\)[/tex], we can use the Rydberg formula for hydrogen:
[tex]\[
\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)
\][/tex]
Where:
- [tex]\(\lambda\)[/tex] is the wavelength of the emitted light,
- [tex]\(R\)[/tex] is the Rydberg constant (approximately [tex]\(1.097373 \times 10^7 \, \text{m}^{-1}\)[/tex]),
- [tex]\(n_i\)[/tex] is the initial energy level (in this case, [tex]\(n_i = 5\)[/tex]),
- [tex]\(n_f\)[/tex] is the final energy level (in this case, [tex]\(n_f = 1\)[/tex]).
Let's plug in the given values:
[tex]\[
R \approx 1.097373 \times 10^7 \, \text{m}^{-1}
\][/tex]
[tex]\[
n_i = 5
\][/tex]
[tex]\[
n_f = 1
\][/tex]
Using these values in the Rydberg formula, the equation becomes:
[tex]\[
\frac{1}{\lambda} = 1.097373 \times 10^7 \left( \frac{1}{1^2} - \frac{1}{5^2} \right)
\][/tex]
First, calculate the terms inside the parentheses:
[tex]\[
\frac{1}{1^2} = 1
\][/tex]
[tex]\[
\frac{1}{5^2} = \frac{1}{25} = 0.04
\][/tex]
So,
[tex]\[
\frac{1}{\lambda} = 1.097373 \times 10^7 \left( 1 - 0.04 \right) = 1.097373 \times 10^7 \times 0.96
\][/tex]
Next, calculate the product:
[tex]\[
\frac{1}{\lambda} \approx 1.05347808 \times 10^7 \, \text{m}^{-1}
\][/tex]
Taking the reciprocal to find [tex]\(\lambda\)[/tex]:
[tex]\[
\lambda \approx \frac{1}{1.05347808 \times 10^7} \approx 9.4923651 \times 10^{-8} \, \text{m}
\][/tex]
Thus, the wavelength of light that is emitted is approximately [tex]\(9.492 \times 10^{-8} \, \text{m}\)[/tex].
Among the given options, the closest value to this result is:
[tex]\[
\boxed{9.50 \times 10^{-8} \, \text{m}}
\][/tex]
