Determine the end (final)value of n in a hydrogen atom transition,if the electron starts in n = 2 and the atom absorbs a photon of light with a frequency of 4.57 × 10¹⁴ Hz.

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Quizplus · Accepted Answer

The energy absorbed by the electron corresponds to the difference in energy levels in the hydrogen atom, as described by the formula for the energy difference $\Delta E = E_f - E_i = -13.6 \text{ eV} \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$, where $n_i$ and $n_f$ are the initial and final principal quantum numbers, respectively. Given the initial state $n_i = 2$ and the frequency of the absorbed photon, we can calculate the energy of the photon using $E = h\nu$, where $h$ is Planck's constant and $\nu$ is the frequency. The energy of the photon will match the energy difference between the initial and final states of the electron in the hydrogen atom. Without doing the exact calculation, the transition from $n = 2$ to $n = 3$ requires less energy compared to transitions to higher energy levels (n = 4, 5, 6, etc.), and the given frequency suggests a transition to the next higher energy level, which is $n = 3$.

Determine the End (Final)value of N in a Hydrogen Atom