Answer to Question #218280 in Physical Chemistry for Mariam

Question

As a result of the transition of an electron between two energy levels light of wavelength 540nm is emitted. Calculate the energy difference between the two levels.

Answer 1

$E = h ⋅ ν$

where

E - the energy of the photon

h - Planck's constant, equal to $6.626×10^{−34}J s$

ν - the frequency of the photon

Now, notice that you are given the wavelength of the photon, λ

As you know, frequency and wavelength have an inverse relationship described by the equation

$λ ⋅ ν = c$

, where

c - the speed of light in vacuum, approximately equal to $3×10^8m s^{−1}$

$λ ⋅ ν = c ⇒ ν = \frac{c} λ$

$E = h ⋅ \frac{c} λ$

This means that the relationship between energy and wavelength looks like this

Another important thing to notice here is that the wavelength of the photon is given in nanometers, nm

. You need to convert this to meters, the unit used for the value of the speed of light.

$E = 6.626 × 10 ^{− 34} J s × \frac{3 × 10^ 8 m s} {540 × 10^ {− 9} m} = 3.681 × 10^ {− 19} J$