In the Lyman series of hydrogen,which are the two maximum wavelengths? Why?
(N/A) The wavelength $\lambda$ for the hydrogen spectrum is given by the Rydberg formula: $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the Lyman series,$n_1 = 1$. The wavelength is inversely proportional to the energy difference $\Delta E = E_{n_2} - E_{n_1}$.
To obtain the maximum wavelengths,we need the minimum energy transitions,which correspond to the smallest values of $n_2$ above $n_1$.
For $n_2 = 2$,$\frac{1}{\lambda_1} = R_H (1 - \frac{1}{4}) = \frac{3}{4} R_H$,giving $\lambda_1 \approx 121.6 \ nm$.
For $n_2 = 3$,$\frac{1}{\lambda_2} = R_H (1 - \frac{1}{9}) = \frac{8}{9} R_H$,giving $\lambda_2 \approx 102.6 \ nm$.
Thus,the two maximum wavelengths correspond to transitions from $n_2 = 2$ and $n_2 = 3$ to $n_1 = 1$.
For the Lyman series,$n_1 = 1$. The wavelength is inversely proportional to the energy difference $\Delta E = E_{n_2} - E_{n_1}$.
To obtain the maximum wavelengths,we need the minimum energy transitions,which correspond to the smallest values of $n_2$ above $n_1$.
For $n_2 = 2$,$\frac{1}{\lambda_1} = R_H (1 - \frac{1}{4}) = \frac{3}{4} R_H$,giving $\lambda_1 \approx 121.6 \ nm$.
For $n_2 = 3$,$\frac{1}{\lambda_2} = R_H (1 - \frac{1}{9}) = \frac{8}{9} R_H$,giving $\lambda_2 \approx 102.6 \ nm$.
Thus,the two maximum wavelengths correspond to transitions from $n_2 = 2$ and $n_2 = 3$ to $n_1 = 1$.
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$(II)$ The integer $n_{1}$ is equal to $2$.
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$(IV)$ The ionization energy of hydrogen can be calculated from wave number of these lines.
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