What is the energy difference between the two quantum states involved in the transition of red light with wavelength 705 nm being absorbed by an atomic gas?

Question

Benjamin Asiedu · Accepted Answer

I found:
#E=h*nu=2.82xx10^-19J=1.76eV# A wavelength of #705 nm# corresponds to a frequency: #nu=c/lambda=(3xx10^8)/(7.05xx10^-7)=4.25xx10^14Hz# According to Einstein's Relativity, the photon energy will be: #color(red)(E=h*nu)=6.63xx10^-34*4.25xx10^14=2.82xx10^-19J=1.76eV# This is precisely the energy gap (or difference) between the two quantum states that your red light photon is absorbing.

Isabella Alvarez · Answer

undefined The energy difference between the two quantum states involved in the transition of red light with a wavelength of 705 nm being absorbed by an atomic gas can be calculated using the formula:

ΔE = hc/λ

Where:
ΔE is the energy difference between the two states,
h is Planck's constant (6.626 x 10^-34 J*s),
c is the speed of light in vacuum (3.00 x 10^8 m/s),
λ is the wavelength of the light.

Plugging in the values:

ΔE = (6.626 x 10^-34 J*s * 3.00 x 10^8 m/s) / (705 x 10^-9 m)

ΔE ≈ 8.88 x 10^-19 Joules