Question

The diagram above represents the energy states for a particular atom. The energies W, X, Y, and Z are all positive and are in Joules.
What is an expression for the second longest wavelength (in meters) that can be emitted by this atom when an electron falls from some higher energy state into the ground state (where $n=1$) in terms of the numbers and/or symbols in the diagram and fundamental constants (like the speed of light, c, and Planck's constant, h)?

• A. $\frac{hc}{2}$
• B. $\frac{cZ}{2h}$
• C. $\frac{hc}{|Y-Z|}$
• D. $\frac{Z}{2hc}$
• E. $\frac{hc}{|X-Z|}$

Answer 1

The correct option is E $\frac{hc}{|X-Z|}$

Wavelength of the photon emitted, $\lambda =\frac{hc}{\Delta E}$ where $\Delta E$ is the energy difference between two states.

For second longest wavelength to be emitted, transition takes place from $n=3$ to $n=1$.

Thus, energy difference between $n=3$ to $n=1$, $\Delta E=|E_3-E_1|=|X-Z|$

$\therefore$ Wavelength of the photon emitted, $\lambda =\frac{hc}{|X-Z|}$