Calculate the wavelength of the emitted radiation when an electron transitions from $n = 3$ to $n = 2$ in a hydrogen atom. To which region does this radiation belong?
(N/A) Using the Rydberg formula for a hydrogen atom: $\frac{1}{\lambda} = R_H \times Z^2 \times (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.
Given $n_1 = 2$,$n_2 = 3$,and $R_H = 1.097 \times 10^7 \ m^{-1}$.
$\frac{1}{\lambda} = 1.097 \times 10^7 \times (\frac{1}{2^2} - \frac{1}{3^2}) = 1.097 \times 10^7 \times (\frac{1}{4} - \frac{1}{9}) = 1.097 \times 10^7 \times \frac{5}{36}$.
$\frac{1}{\lambda} = 1.5236 \times 10^6 \ m^{-1}$.
$\lambda = 6.563 \times 10^{-7} \ m = 656.3 \ nm$.
This wavelength corresponds to the Balmer series,which lies in the visible region.
Given $n_1 = 2$,$n_2 = 3$,and $R_H = 1.097 \times 10^7 \ m^{-1}$.
$\frac{1}{\lambda} = 1.097 \times 10^7 \times (\frac{1}{2^2} - \frac{1}{3^2}) = 1.097 \times 10^7 \times (\frac{1}{4} - \frac{1}{9}) = 1.097 \times 10^7 \times \frac{5}{36}$.
$\frac{1}{\lambda} = 1.5236 \times 10^6 \ m^{-1}$.
$\lambda = 6.563 \times 10^{-7} \ m = 656.3 \ nm$.
This wavelength corresponds to the Balmer series,which lies in the visible region.
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