In a transition from one vibrational state to another, a molecule emits a photon of wavelength $5.56$ $\mu \mathrm { m }$. What is the energy difference between these two states? $\left( c = 3.00 \times 10 ^ { 8 } \mathrm {~m} / \mathrm { s } , h = 6.626 \times 10 ^ { - 34 } \mathrm {~J} \right.$ $\cdot \mathrm { s } , 1 \mathrm { eV } = 1.60 \times 10 ^ { - 19 } \mathrm {~J}$ )
A) $2.23 \mathrm { MeV }$
B) $0.223 \mathrm { eV }$
C) $0.223 \mathrm { MeV }$
D) $13.6 \mathrm { MeV }$
E) $13.6 \mathrm { eV }$

Question

Quizplus · Accepted Answer

The energy difference between two states can be calculated using the formula $E = \frac{hc}{\lambda}$, where $E$ is the energy in joules, $h$ is Planck's constant ($6.626 \times 10^{-34} \mathrm{~J \cdot s}$), $c$ is the speed of light ($3.00 \times 10^8 \mathrm{~m/s}$), and $\lambda$ is the wavelength of the emitted photon ($5.56 \times 10^{-6} \mathrm{~m}$). Substituting the given values, we find $E$ in joules, and then convert it to electron volts (eV) using the conversion factor $1 \mathrm{eV} = 1.60 \times 10^{-19} \mathrm{~J}$. The calculation yields an energy difference of approximately $0.223 \mathrm{~eV}$.

In a Transition from One Vibrational State to Another, a Molecule