Find the ionization energy and the wavelength of the photon emitted for a hydrogen atom transition from $n_1 = 2$ to $n_2 = \infty$.
(A) The energy required to remove an electron from the $n = 2$ state to infinity is given by $E = R_H \times h \times c \times (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.
Substituting $n_1 = 2$ and $n_2 = \infty$,we get $E = 2.18 \times 10^{-18} \ J \times (\frac{1}{2^2} - 0) = 5.45 \times 10^{-19} \ J$.
The wavelength $\lambda$ is calculated using $\lambda = \frac{hc}{E}$.
$\lambda = \frac{6.626 \times 10^{-34} \ J \cdot s \times 3 \times 10^8 \ m/s}{5.45 \times 10^{-19} \ J} = 3.64 \times 10^{-7} \ m$.
Substituting $n_1 = 2$ and $n_2 = \infty$,we get $E = 2.18 \times 10^{-18} \ J \times (\frac{1}{2^2} - 0) = 5.45 \times 10^{-19} \ J$.
The wavelength $\lambda$ is calculated using $\lambda = \frac{hc}{E}$.
$\lambda = \frac{6.626 \times 10^{-34} \ J \cdot s \times 3 \times 10^8 \ m/s}{5.45 \times 10^{-19} \ J} = 3.64 \times 10^{-7} \ m$.
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