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

`4.98\cdot 10^(-19) J`

Step-by-step explanation:

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

`E=(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=((6.63\cdot 10^(-34))(3\cdot 10^8))/(400\cdot 10^(-9))=4.98\cdot 10^(-19) J`