Question

Silver emits light of 328.1 nm when burned. What is the energy of a mole of these photons?

Answer 1

Just under 4 eV

Explanation:

You can use the relationship `E = h.nu` where `E` is the energy, `h` is Planck's constant and `nu` is the frequency.

However, we do not have frequency, we only have wavelength. So we need to express frequency as follows:

`c = nu.lambda`

`nu = c/lambda`

Where `c` is the speed of light, `lambda` is the wavelength and `nu` is the frequency.

Therefore you can now say: `E = (h.c)/lambda`

But you need to ensure that your units are consistent. Units of Planck's constant are `J.s`, speed of light is commonly quoted in m/s, so thats fine. `nu` is in Hertz which is units of `s^-1`. So again, that ties in OK.

Wavelength, however, is given in nm so in order to ensure that everything balances, you need to express this as metres, (328.1 x `10^-9` m).

Now you can work it out:

`E = (h.c)/lambda`

`= (6.626 times 10^-34. 3 times 10^8)/(328.1 times 10^-9)`

= `6.0585 times 10^-19` J

However, as this is a very small amount of energy, it is probably better to express it in units of "electron volts" - one electron volt is `1.602 times 10^-19` J, in which case the answer works out to be 6.0585/1.602 = 3.78 eV.

(One electron volt is the energy involved in the charge of a single electron moving through a potential difference of one volt, and it is a useful unit for expressing extremely small amounts of energy).