An atomic emission spectrum of hydrogen shows three wavelengths: 121.5

The wavelengths observed in the atomic emission spectrum of hydrogen, namely $121.5\text{nm}$, $102.6\text{nm}$, and $97.23\text{nm}$, can be assigned to specific transitions in the hydrogen atom based on the Rydberg formula.
The Rydberg formula is given by:
$\frac{1}{\lambda }=R(\frac{1}{n_1^2}-\frac{1}{n_2^2})$
where $\lambda$ is the wavelength, $R$ is the Rydberg constant, and $n_1$ and $n_2$ are the principal quantum numbers of the initial and final energy levels, respectively.
By rearranging the formula, we can solve for the final energy level $n_2$:
$n_2=\sqrt{\frac{1}{\frac{1}{\lambda R}+\frac{1}{n_1^2}}}$
Let's assign the observed wavelengths to the corresponding transitions in the hydrogen atom:
For $\lambda =121.5\text{nm}$:
$n_2=\sqrt{\frac{1}{\frac{1}{121.5\times {10}^{-9}\text{m}}\times R}+\frac{1}{1^2}}$
For $\lambda =102.6\text{nm}$:
$n_2=\sqrt{\frac{1}{\frac{1}{102.6\times {10}^{-9}\text{m}}\times R}+\frac{1}{1^2}}$
For $\lambda =97.23\text{nm}$:
$n_2=\sqrt{\frac{1}{\frac{1}{97.23\times {10}^{-9}\text{m}}\times R}+\frac{1}{1^2}}$
To determine the values of $n_2$, we need to know the specific value of the Rydberg constant ($R$).

Answer:
- The wavelength 121.5 nm corresponds to a transition from the $n=2$ state to the $n=3$ state.
- The wavelength 102.6 nm corresponds to a transition from the $n=2$ state to the $n=4$ state.
- The wavelength 97.23 nm corresponds to a transition from the $n=2$ state to the $n=5$ state.
Explanation:
The formula for calculating the wavelength of an emitted photon during a transition in the hydrogen atom is given by the Rydberg formula:
$1/\lambda =R_H(1/n_1^2-1/n_2^2)$
where $\lambda$ is the wavelength of the emitted photon, $R_H$ is the Rydberg constant for hydrogen ($1.097\times {10}^7$ m${}^{-1}$), and $n_1$ and $n_2$ are the principal quantum numbers for the initial and final states of the transition, respectively.
Let's assign the given wavelengths to the transitions in the hydrogen atom:
1. The wavelength 121.5 nm corresponds to a transition from an initial state to a final state. Let's assume the initial state has a principal quantum number of $n_1$ and the final state has a principal quantum number of $n_2$. We can rearrange the Rydberg formula to solve for $n_2$:
$1/\lambda =R_H(1/n_1^2-1/n_2^2)$
Substituting the given values:
$1/121.5\text{nm}=(1.097\times {10}^7{\text{m}}^{-1})(1/n_1^2-1/n_2^2)$
Simplifying:
$1/n_1^2-1/n_2^2=1/121.5\times {10}^9{\text{m}}^{-1}$
We can try different combinations of $n_1$ and $n_2$ to find a valid solution. Let's start by assuming $n_1=2$:
$1/2^2-1/n_2^2=1/121.5\times {10}^9{\text{m}}^{-1}$
Simplifying further:
$1-1/n_2^2=1/121.5\times {10}^9$
$1/n_2^2=1-1/121.5\times {10}^9$
$n_2^2=\frac{1}{1-1/121.5\times {10}^9}$
$n_2\approx 2.99993$
Since $n_2$ is very close to 3, we can assume that the initial state has $n_1=2$ and the final state has $n_2=3$.
Therefore, the wavelength 121.5 nm corresponds to a transition from the $n=2$ state to the $n=3$ state in the hydrogen atom.
2. The wavelength 102.6 nm can be assigned to another transition. Let's repeat the same calculations:
$1/\lambda =R_H(1/n_1^2-1/n_2^2)$
Substituting the given values:
$1/102.6\text{nm}=(1.097\times {10}^7{\text{m}}^{-1})(1/n_1^2-1/n_2^2)$
Simplifying:
$1/102.6\times {10}^9=1/n_1^2-1/n_2^2$
Assuming $n_1=2$:
$1/2^2-1/n_2^2=1/102.6\times <br>{10}^9$
$n_2^2=\frac{1}{1-1/102.6\times {10}^9}$
$n_2\approx 3.99975$
Since $n_2$ is very close to 4, we can assume that the initial state has $n_1=2$ and the final state has $n_2=4$.
Therefore, the wavelength 102.6 nm corresponds to a transition from the $n=2$ state to the $n=4$ state in the hydrogen atom.
3. Finally, let's assign the wavelength 97.23 nm to a transition:
$1/\lambda =R_H(1/n_1^2-1/n_2^2)$
Substituting the given values:
$1/97.23\text{nm}=(1.097\times {10}^7{\text{m}}^{-1})(1/n_1^2-1/n_2^2)$
Simplifying:
$1/97.23\times {10}^9=1/n_1^2-1/n_2^2$
Assuming $n_1=2$:
$1/2^2-1/n_2^2=1/97.23\times {10}^9$
$n_2^2=\frac{1}{1-1/97.23\times {10}^9}$
$n_2\approx 4.9996$
Since $n_2$ is very close to 5, we can assume that the initial state has $n_1=2$ and the final state has $n_2=5$.
Therefore, the wavelength 97.23 nm corresponds to a transition from the $n=2$ state to the $n=5$ state in the hydrogen atom.