A hydrogen atom is excited to the $n = 9$ level. Its decay to the $n = 6$ level detected in a photographic plate. What is the frequency of the light photographed? $\left( 1 \mathrm { eV } = 1.60 \times 10 ^ { - 19 } \mathrm {~J} , h = 6.626 \times 10 ^ { - 34 } \mathrm {~J} \right.$. s)
A) $5910 \mathrm {~Hz}$
B) $3.28 \times 10 ^ { - 9 } \mathrm {~km}$
C) $5.08 \times 10 ^ { 13 } \mathrm {~Hz}$
D) $5910 \mathrm {~nm}$

Question

Quizplus · Accepted Answer

The energy difference between two levels in a hydrogen atom is given by the formula $\Delta E = -13.6 \, \text{eV} \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$, where $n_i$ and $n_f$ are the initial and final principal quantum numbers, respectively. For a transition from $n = 9$ to $n = 6$, we calculate the energy difference and then use the relation $E = h\nu$ to find the frequency $\nu$. Given that $1 \, \text{eV} = 1.60 \times 10^{-19} \, \text{J}$ and $h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s}$, we can calculate the frequency of the emitted light, which results in a value in the order of $10^{13} \, \text{Hz}$, making option C correct.

A Hydrogen Atom Is Excited to The n=9n = 9n=9 Level Its Decay to The to The n

A Hydrogen Atom Is Excited to The $n = 9$ Level Its Decay to The to The n