Question

We say that visible light has wavelength from 400nm to roughly 800nm. What is the biggest "energy-jump" (excited) for an atom if the photon which was sent out was visible? What's the biggest "energy-jump"?
(Give the answer in joules)

thanks !! ;-)

Answer 1

Answer:

$4.98\cdot 10^{-19} J$

Explanation:

The energy of the emitted photon is inversely proportional to its wavelength, according to the equation:

$E=\frac{hc}{\lambda}$

where

$h=6.63\cdot 10^{-34} Js$ is the Planck's constant

$c=3.0\cdot 10^8 m/s$ is the speed of light

$\lambda$ is the wavelength

This means that the biggest energy is released when the wavelength is the shortest. For a photon of visible light, the shortest wavelength is

$\lambda=400 nm = 400\cdot 10^{-9} m$

So, substituting into the equation, we find the corresponding energy:

$E=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{400\cdot 10^{-9}}=4.98\cdot 10^{-19} J$