# Question #42282

Apr 30, 2016

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 $1.6 \times {10}^{-} 19 J .$

So $2.11 e V = 2.11 \times 1.6 \times {10}^{-} 19 = 3.376 \times {10}^{-} 19 J$

The energy of a photon is give by:

$E = h f = h \frac{c}{l} a m \mathrm{da}$
Where h = Planck's constant = $6.63 \times {10}^{-} 34 J {s}^{-} 1$; and
c = speed of light= $3 \times {10}^{8} m {s}^{-} 1$

Hence we can solve for the photon in question:

$3.376 \times {10}^{-} 19 = \left(6.63 \times {10}^{-} 34\right) \cdot \frac{3 \times {10}^{8}}{l} a m \mathrm{da}$
$l a m \mathrm{da} = 5.89 \times {10}^{-} 7 m = 589 n m$.

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