Question

Your eyes have three different types of cones with maximum absorption at 437 nm, 533 nm, and 564 nm.What photon energies correspond to these wavelengths? (answer in eV)

Answer 1

Answers:

The energy
`E` of a photon is given by the following formula:

`E=h.f` (1)

Where:

`h=4.136(10)^(-15) eV.s` is the Planck constant

`f` is the frequency in hertz
`Hz=s^(-1)`

Now, the frequency has an inverse relation with the wavelength
`\lambda`:

`f=(c)/(\lambda)` (2)

Where
`c=3(10)^(8)m/s` is the speed of light in vacuum

Substituting (2) in (1):

`E=(hc)/(\lambda)` (3)

Knowing this, let's begin with the answers:

437 nm

For
`\lambda=437nm=437(10)^(-9)m`

`E=((4.136(10)^(-15) eV.s)(3(10)^(8)m/s))/(437(10)^(-9)m)`

`E=(1.24(10)^(-6)eV.m )/(437(10)^(-9)m)`

`E=2.837eV`

533 nm

For
`\lambda=533nm=533(10)^(-9)m`

`E=(1.24(10)^(-6)eV.m )/(533(10)^(-9)m)`

`E=2.327eV`

564 nm

For
`\lambda=564nm=564(10)^(-9)m`

`E=(1.24(10)^(-6)eV.m )/(564(10)^(-9)m)`

`E=2.2eV`