Answer to Question #138598 in Electric Circuits for Mac Evans

Question

The Balmer series for the hydrogen atom corresponds to electronic transitions that terminate in the state with quantum number n=2. Consider the photon of longest wavelength corresponding to a transition. Determine (a) its energy and (b) its wavelength.

Answer 1

Since we need to consider the photon with the longest wavelength that ends the electron transition in the state with the quantum number $n=2$ , this will be transition $3-2$ since the photon energy formula $E=h\times \frac{c}{\lambda}$ , (where $h$ is the Planck constant, $c$ is the speed of light, and $\lambda$ is the wavelength) shows that the photon energy is inversely proportional to the wavelength, and since a larger transition corresponds to a large absorption/release of energy, we choose the smallest transition (transition $1-2>$ transitions $2-3$ ). By the formula $\frac{1}{\lambda}=R\times(\frac{1}{m^2}-\frac{1}{n^2})$ (where $R$ is the Rydberg constant, and $n$ and $m$ are the quantum numbers of the photon state) $\frac{1}{\lambda}=3.29\times10^{15}\times(\frac{1}{2^2}-\frac{1}{3^2})$ $=\frac{5\times 3.29\times10^{15}}{36}$ therefore $\lambda=\frac{36}{5\times 3.29\times10^{15}}=2.19\times 10^{-15}$ m.

$E=6.63\times 10^{-34}\times \frac{3\times 10^8}{2.19\times 10^{-15} }=9.08\times 10^{-11}$ J.