Question

A photographic film is coated with a silver bromide layer. When light falls on this film, silver bromide molecules dissociate and the film records the light there. A minimum of 0.6 eV is needed to dissociate a silver bromide molecule. Find the maximum wavelength of light that can be recorded by the film.

Answer 1

Wavelength is the length of something very seriously

Answer 2

The maximum wavelength of light that can be recorded by the film is 2071 nm

Explanation:

Given data in the question  

Function of Work,   $W_{0}=0.6 \mathrm{eV}$

Now, Function of work,  $W_{0}=\frac{h c}{\lambda}$

$where,\\ h= Planck's constant\\ \lambda= light wavelength\\ c= light of velocity$

$\therefore \lambda=\frac{h c}{w_{0}}$

Step 1:

$=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{0.6 \times 1.6 \times 10^{-19}}$

$=\frac{19.89 \times 10^{-34} \times 10^{8}}{0.6 \times 1.6 \times 10^{-19}}$

$=\frac{19.89 \times 10^{-34} \times 10^{8}}{0.96 \times 10^{-19}}$

Step 2:

$=\frac{19.89 \times 10^{-26}}{0.96 \times 10^{-19}}$

$=\frac{19.89 \times 10^{-7}}{0.96}$

$=20.71 \times 10^{-7} \mathrm{m}$

=2071 nm