The wavelength of the emitted photon if an electron in the hydrogen atom makes a transition from the n = 2 state to the ground state is closest to which of the following values? (c = 3.0 × 10⁸ m/s,h = 6.626 × 10^-34 J ∙ s,1 eV = 1.60 × 10^-19J) A)122 nm B)203 nm C)243 nm D)389 nm E)411 nm

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Quizplus · Accepted Answer

The energy difference between the n = 2 state and the ground state (n = 1) in a hydrogen atom can be calculated using the formula for the energy levels of a hydrogen atom, $E_n = -\frac{13.6 \text{ eV}}{n^2}$. The energy difference ($\Delta E$) is then $\Delta E = E_2 - E_1 = -3.4 \text{ eV} - (-13.6 \text{ eV}) = 10.2 \text{ eV}$. Converting this energy into joules ($1 \text{ eV} = 1.60 \times 10^{-19} \text{ J}$), we get $\Delta E = 10.2 \text{ eV} \times 1.60 \times 10^{-19} \text{ J/eV} = 1.632 \times 10^{-18} \text{ J}$. The wavelength ($\lambda$) of the emitted photon can be found using the equation $E = \frac{hc}{\lambda}$, solving for $\lambda$ gives $\lambda = \frac{hc}{E}$. Substituting $h = 6.626 \times 10^{-34} \text{ J} \cdot \text{s}$, $c = 3.0 \times 10^8 \text{ m/s}$, and $E = 1.632 \times 10^{-18} \text{ J}$, we find $\lambda \approx 122 \text{ nm}$.

The Wavelength of the Emitted Photon If an Electron in the Hydrogen