What frequency must a photon have to raise an electron in a hydrogen atom from the n = 2 to the n = 4 state? (h = 6.626 × 10^-34 J ∙ s,1 eV = 1.60 × 10^-19J) A)3.06 × 10¹⁴ Hz B)6.16 × 10¹⁴ Hz C)4.11 × 10¹⁴ Hz D)5.20 × 10¹⁴ Hz E)9.24 × 10¹⁴ Hz

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Quizplus · Accepted Answer

The energy difference between the n = 2 and n = 4 levels in a hydrogen atom can be calculated using the formula for the energy levels of a hydrogen atom, $E_n = -\frac{13.6 \text{ eV}}{n^2}$, where $n$ is the principal quantum number. The energy difference ($\Delta E$) is then given by $\Delta E = E_4 - E_2$. Plugging in the values, we find $\Delta E = -\frac{13.6 \text{ eV}}{4^2} - (-\frac{13.6 \text{ eV}}{2^2})$, which simplifies to $\Delta E = 2.55 \text{ eV}$. Converting this energy difference to joules ($1 \text{ eV} = 1.60 \times 10^{-19} \text{ J}$), we get $\Delta E = 2.55 \text{ eV} \times 1.60 \times 10^{-19} \text{ J/eV} = 4.08 \times 10^{-19} \text{ J}$. The frequency ($f$) of the photon can then be found using the equation $E = hf$, where $h$ is Planck's constant ($6.626 \times 10^{-34} \text{ J} \cdot \text{s}$). Solving for $f$, we get $f = \frac{E}{h} = \frac{4.08 \times 10^{-19} \text{ J}}{6.626 \times 10^{-34} \text{ J} \cdot \text{s}} \approx 6.16 \times 10^{14} \text{ Hz}$.

What Frequency Must a Photon Have to Raise an Electron