A hydrogen atom is excited to the n = 11 level.Its decay to the n = 7 level is detected in a photographic plate.What is the wavelength of the light photographed? (c = 3.00 × 10⁸ m/s,1 eV = 1.60 × 10^-19J,h = 6.626 × 10^-34 J ∙ s) A)7,510 nm B)4,670 nm C)12,400 nm D)4,380 nm

Question

Quizplus · Accepted Answer

The energy difference between the n = 11 and n = 7 levels can be calculated using the Balmer-Rydberg equation:
$\frac{1}{\lambda}=R_H(\frac{1}{n_1^{2}}-\frac{1}{n_2^{2}})$
where $R_H=1.096776	imes10^7 m^{-1}$ is the Rydberg constant for hydrogen, $n_1=7$, and $n_2=11$. Solving for $\lambda$, we get:
$\lambda= 7510$ nm. 
Therefore, the correct answer is choice (A).

A Hydrogen Atom Is Excited to the N = 11