In the He-Ne laser, the lasing transition goes from the metastable 20.66 eV state in Ne to a lower-energy state. The wavelength of the laser photons is 633.0 nm. What is the energy of the lower state of the lasing transition?
A) 0
B) 4.660 eV
C) 13.61 eV
D) 18.70 eV
E) 20.61 eV

Question

Quizplus · Accepted Answer

The energy of the laser photons can be calculated using the formula $E = \frac{hc}{\lambda}$, where $h$ is Planck's constant ($6.626 \times 10^{-34} \, \text{J s}$), $c$ is the speed of light ($3.00 \times 10^8 \, \text{m/s}$), and $\lambda$ is the wavelength of the light ($633.0 \, \text{nm} = 633.0 \times 10^{-9} \, \text{m}$). The energy of the emitted photon is $E = \frac{(6.626 \times 10^{-34} \, \text{J s})(3.00 \times 10^8 \, \text{m/s})}{633.0 \times 10^{-9} \, \text{m}}$, which gives approximately $3.14 \times 10^{-19} \, \text{J}$. Converting this energy to electron volts ($1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}$), we get approximately $1.96 \, \text{eV}$. Since the transition starts from the 20.66 eV state and emits a photon of approximately 1.96 eV, the energy of the lower state is $20.66 \, \text{eV} - 1.96 \, \text{eV} \approx 18.70 \, \text{eV}$.

In the He-Ne Laser, the Lasing Transition Goes from the Metastable