# 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?

Jun 22, 2015

I found:
$E = h \cdot \nu = 2.82 \times {10}^{-} 19 J = 1.76 e V$

#### Explanation:

A wavelength of $705 n m$ corresponds to a frequency:
$\nu = \frac{c}{\lambda} = \frac{3 \times {10}^{8}}{7.05 \times {10}^{-} 7} = 4.25 \times {10}^{14} H z$

From Einstein's Relatonship you have that the photon energy will be:

$\textcolor{red}{E = h \cdot \nu} = 6.63 \times {10}^{-} 34 \cdot 4.25 \times {10}^{14} = 2.82 \times {10}^{-} 19 J = 1.76 e V$

This is exactly the energy gap (or difference) between the two quantum states involved in the absorption of your photon of red light.