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

Jan 15, 2018

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:

$c = \nu . \lambda$

$\nu = \frac{c}{\lambda}$

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

Therefore you can now say: $E = \frac{h . c}{\lambda}$

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:

$E = \frac{h . c}{\lambda}$

$= \frac{6.626 \times {10}^{-} 34. 3 \times {10}^{8}}{328.1 \times {10}^{-} 9}$

= $6.0585 \times {10}^{-} 19$ 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).