Question

The energy in electron-volts) for a photon with awavelength 724.4 nm is:

Answer 1

Use the Planck expression to find the energy of a photon.

`E=hf`

Where h = 6.63x10^-34 J*s and f = c/lambda.

`\begin{gathered} E=6.63*10^(-34)J\cdot s\cdot(c)/(\lambda)=6.63*10^(-34)J\cdot s\cdot(3*10^8\cdot(m)/(s))/(724.4*10^(-9)m) \\ E=2.75*10^(-19)J \end{gathered}`

Now, transform the energy from Jules to electron-volts. We know that 1 electron-volt equals 1.60x10^-9 J.

`E=2.75*10^(-19)J\cdot(1eV)/(1.6*10^(-19)J)=1.72eV`

Therefore, the energy of the photon with the given wavelength is 1.72 eV.