15.2 Emission and absorption spectra  (Page 3/4)

We are given $\lambda =500\times {10}^{-9}\mathrm{m}$ and we need to find the photon's energy , frequency and colour .

The energy of the photon is $3,98\times {10}^{-19}$ J.

The frequency of the photon is $6\times {10}^{14}\mathrm{Hz}$ .

The wavelength given in the question is 500 nm. We can see in the table that green light has wavelengths between 492 - 577 nm. Therefore 500 nm is in this range so the colour of the light is green .

Four wavelengths of light are given and we need to find their colours .

We also need to find which colour photon has the highest energy and which one has the lowest energy.

1. 400 nm falls into the range for violet light (390 - 455 nm).
2. 580 nm falls into the range for yellow light (577 - 597 nm).
3. 650 nm falls into the range for red light (622 - 780 nm).
4. 300 nm is not shown in the table. However, this wavelength is just a little shorter than the shortest wavelength in the violet range. Therefore 300 nm is ultraviolet .

We know $E=\frac{hc}{\lambda }$

For 400 nm:

For 580 nm:

For 650 nm:

For 300 nm:

Therefore, the photons with the highest energy are the ultraviolet photons.

The photons with the lowest energy are from light which is red .