Question

Question #42282

Answer 1

The photon is on the border between yellow and orange

Explanation:

To determine the colour, we will need to know its wavelength (or frequency). We can then compare this to the known wavelengths of colour.

The question tells us the photon has 2.11 eV which is a unit of energy. In the first case, let's transfer this into joules. The energy of 1eV is defined as the energy transferred to or from an electron when it moves through a potential difference of 1 volt. This is equivalent to the electron's charge times the pd of 1v, which equals `1.6 times 10^-19J.`

So `2.11 eV = 2.11times1.6times 10^-19=3.376 times 10^-19J`

The energy of a photon is give by:

`E=hf=hc/lamda`
Where h = Planck's constant = `6.63 times 10^-34Js^-1`; and
c = speed of light= `3 times 10^8 ms^-1`

Hence we can solve for the photon in question:

`3.376 times 10^-19=(6.63times 10^-34)*(3times 10^8)/lamda`
`lamda=5.89times 10^-7m=589nm`.

If we look at a diagram of the visible electromagnetic spectrum, we see that this is on the border between yellow and orange.