An energy of 13.6 eV is needed to remove an electron from the n = 3 state of a lithium atom.If a single photon accomplishes this task,what wavelength is needed? (h = 6.63 × 10−³⁴ J⋅s,c = 3.00 × 10⁸ m/s,1 eV = 1.6 × 10−¹⁹ J,and 1 nm = 10−⁹ m) A)146.3 nm B)30.5 nm C)13.8 nm D)91.4 nm E)234 nm

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Quizplus · Accepted Answer

To find the wavelength of the photon, we use the energy-wavelength relationship: $E = \frac{hc}{\lambda}$, where $E$ is the energy in joules, $h$ is Planck's constant ($6.63 \times 10^{-34} \, \text{J} \cdot \text{s}$), $c$ is the speed of light ($3.00 \times 10^8 \, \text{m/s}$), and $\lambda$ is the wavelength in meters. First, convert the energy from eV to joules: $13.6 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 2.176 \times 10^{-18} \, \text{J}$. Then, rearrange the formula to solve for $\lambda$: $\lambda = \frac{hc}{E}$. Substituting the values, $\lambda = \frac{6.63 \times 10^{-34} \, \text{J} \cdot \text{s} \times 3.00 \times 10^8 \, \text{m/s}}{2.176 \times 10^{-18} \, \text{J}} \approx 9.14 \times 10^{-8} \, \text{m} = 91.4 \, \text{nm}$.

An Energy of 13