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 0.6 eV, the smallest amount of energy needed to dissociate an 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? ​

Answer 1

Hint: in order to solve the question, we will use the relation between frequency and energy and after that we will derive the relation between energy and wavelength to solve the first part and for the second we will use the electromagnetic spectrum region table and solve the question.

Formula used:
$E = hf$
Here, $E$ refers to energy, $h$ refers to planck's constant and $f$ refers to frequency.
$f = \dfrac{c}{\lambda }$
Here, $c$ refers to speed of light, $\lambda $ refers to wavelength and $f$ refers to frequency.

Complete step by step answer:
It is given in the question that photon absorption will occur if the photon energy equals or exceeds 0.6 eV and we require the, the smallest amount of energy needed to dissociate an AgBr molecule in the film. The smallest amount of energy which will dissociate AgBr molecules in the film is 0.6 ev.

(a) The energy of the photon is equal to the frequency of the photon multiplied by the planck's constant that can be written in the equation as
$E = hf$ ……..(1)
and the frequency can be written as
$f = \dfrac{c}{\lambda }$ ……… (2)
Now by using the equation 1 and 2 we can derive the equation which suggest the relation between energy and frequency
Putting the equation of f into hf
$E = \dfrac{{hc}}{\lambda }h \\
\Rightarrow E= 4.136 \times {10^{ - 15}}ev$
To find the greatest wavelength of light that can be recorded by the film we will substitute the value of $h$ ,$c$ and $E$.
$E = 0.6{\text{ }}ev$
Here, $c$ is the speed of light, it is constant and its value is $c = 3 \times {10^8}\,m{s^{ - 1}}$ and $h$ is planck constant and it’s values in ev unit is $h = 4.136 \times {10^{ - 15}}ev{(Hz)^{ - 1}}$.

Now substituting the values
$E = \dfrac{{hc}}{\lambda }$
$0.6{\text{ }}ev = \dfrac{{4.136 \times {{10}^{ - 15}}ev{{(Hz)}^{ - 1}} \times 3 \times {{10}^8}m{s^{ - 1}}}}{\lambda }$
Now taking $\lambda $ on the other side we get
\[\lambda = \dfrac{{4.136 \times {{10}^{ - 15}}ev{{(Hz)}^{ - 1}} \times 3 \times {{10}^8}m{s^{ - 1}}}}{{0.6{\text{ }}ev}}\]
\[ \Rightarrow \lambda = \dfrac{{4.136 \times 3 \times {{10}^{8 - 15}}}}{{0.6{\text{ }}}}m\]
\[ \therefore \lambda = 2.1 \times {10^{ - 6}}m\]
Hence, \[2.1 \times {10^{ - 6}}m\] greatest wavelength of light that can be recorded by the film.

(b) \[2.1 \times {10^{ - 6}}m\] wavelength in meters. To know the electromagnetic spectrum we will convert it in cm.
$1 m = 100 cm$
Therefore the wavelength will be \[2.1 \times {10^{ - 4}}cm\]. According to the electromagnetic spectrum, this wavelength lies in the infrared region.

Note: Many of the students will make the mistake by taking planck's constant in other unit but in this question we have to take it in the unit which is related to electron volt as all the other units are in electrovolt this mistake can make whole question incorrect along with this we have to take by looking in region table that in which unit wavelength is given as here it is centimeter but answer was in meter so we have to change it.