Question

Monochromatic light (that is, light of a single wavelength) is to be absorbed by a sheet of photographic film and thus recorded on the film. Photon absorption will occur if the photon energy equals or exceeds $\mathbf{0}\mathbf{.}\mathbf{6}\mathbf{}\mathbf{eV}$, the smallest amount of energy needed to dissociate an $\mathbf{AgBr}$ molecule in the film. (a) What is the greatest wavelength of light that can be recorded by the film? (b) In what region of the electromagnetic spectrum is this wavelength located?

Short answer

1. The greatest wavelength of light that can be recorded by the film is $2.1\mathrm{\mu m}$.
2. The region of the electromagnetic spectrum is infrared.

Step-by-step solution

Find the expression for the greatest wavelength of light

The expression for frequency of the monochromatic wave is given by,

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

Here, localid="1663053548195" $f$is the frequency, localid="1663053597284" $\lambda$is the wavelength, localid="1663053648806" $c$is the speed of the light.

The expression for energy of the photon is given by,

localid="1663053690989" $E=hf$

Here, localid="1663053732860" $h$is the Planck’s constant.

Combine the above two equations.

localid="1663053778641" $E=\frac{hc}{\lambda }$

The minimum energy corresponds to the greatest wavelength recorded by the film is given by,

localid="1663053823945" $E_{min}=\frac{hc}{{\lambda }_{greatest}}$

localid="1663053872627" ${\lambda }_{greatest}=\frac{hc}{E_{min}}$ ……. (1)

Determine the greatest wavelength of light

(a)

Substitute $0.6\mathrm{eV}$for $E_{\min },6.626\times {10}^{-34}\mathrm{J}·\mathrm{s}$, for $h$, and $3\times {10}^8$for $c$in equation (1).

$$\begin{gathered} {\lambda }_{\mathrm{greatest}}=\frac{\left(6.626\times {10}^{-34}\mathrm{J}\right)\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)}{\left(0.6\mathrm{eV}\right)\left({\displaystyle \frac{1.6\times {10}^{-19}\mathrm{J}}{1\mathrm{eV}}}\right)} \\ =2.1\times {10}^{-6}\mathrm{m}\left(\frac{1\mathrm{\mu m}}{{10}^{-6}\mathrm{m}}\right) \\ =2.1\mathrm{\mu m} \end{gathered}$$

Therefore, the greatest wavelength of light that can be recorded by the film is $2.1\mathrm{\mu m}$.

Determine the region of the electromagnetic spectrum 

(b)

The infrared region in the electromagnetic spectrum ranges from $0.7\mathrm{\mu m}$to 1mm.

The wavelength $2.1\mathrm{\mu m}$is in the range of $0.7\mathrm{\mu m}$to 1mm. So, the wavelength is located in the infrared region of the electromagnetic spectrum.

Therefore, the region of the electromagnetic spectrum is infrared.