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

Here $W_0=0.6eV$

For $W_0$ to be min '$\lambda$' becomes maximum

$W_0=\frac{hc}{\lambda }$

$\text{or}\lambda =\frac{hc}{W_0}$

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

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

$=2071\text{nm}$