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

Question

Quizplus · Accepted Answer

Since the lasing transition goes from the 20.66 eV state in Ne to a lower state and the laser photons have a wavelength of 633.0 nm, we can use the formula E = hc/λ, where E is the energy of the photon, h is Planck's constant, c is the speed of light, and λ is the wavelength of the photon. If we plug in the given values, we get:

E = (6.626 × 10^-34 J s) × (2.998 × 10^8 m/s) / (633.0 × 10^-9 m) = 3.14 × 10^-19 J

Since the lasing transition goes from the 20.66 eV state in Ne to a lower state, we know that the energy of the photon (3.14 × 10^-19 J) corresponds to the energy difference between the 20.66 eV state in Ne and a lower state. The only answer choice that corresponds to this energy difference is D, the 18.70 eV excited state of Ne. Therefore, the lower state of the lasing transition is the 18.70 eV excited state of Ne.

In the He-Ne Laser,the Lasing Transition Goes from the 20