Question

To expose photographic films. photons of light dissociate silver bromide $\mathit{A}\mathit{g}\mathit{B}\mathit{r}$molecules which requires an energy of $\mathbf{1}\mathbf{.}\mathbf{2}\mathit{e}\mathit{V}$ . What limit does this impose on the wavelengths that may be recorded by photographic film?

Short answer

The wavelengths that may be recorded by photographic film is$1.037\times {10}^{-6}m$

Step-by-step solution

Step 1:

In specifically, a photon's energy is equal to the radiation frequency multiplied by Planck's constant.

The energy of the photon is given by expression,

$$\begin{gathered} E=HF \\ E=\frac{hc}{\mathrm{\pi }} \end{gathered}$$

Here, $\mathit{E}$ is the energy of a photo, $\mathit{h}$is the plank’s constant, $\mathit{c}$is the speed of light, $\mathit{\lambda }$is the wavelength, and$\mathit{f}$ is the frequency.

Calculate the wavelength.

Consider the given data as below.

The energy, $E=1.2eV$

Plank’s constant, $h=6.636\times {10}^{{}^{-34}}J\cdot s$

Speed of light, role="math" localid="1657556142389" $c=3\times {10}^8\frac{\mathrm{m}}{\mathrm{s}}$

To calculate the wavelength rearrange equation (1) as below.

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

Substituting the known parameter to calculate the value of wavelength.

$$\begin{gathered} \lambda =\frac{6.636\times {10}^{-34}\mathrm{J}·\mathrm{s}\times 3\times {10}^8\mathrm{m}/\mathrm{s}}{1.2\mathrm{eV}}\left(\frac{1\mathrm{eV}}{1.6\times {10}^{-19}\mathrm{J}}\right) \\ =1.037\times {10}^{-6}m \end{gathered}$$

Therefore, the limit for the wavelength is respectively $1.037\times {10}^{-6}m$