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 e V is needed to dissociate a silver bromide molecule. Find the maximum wavelength of light that can be recorded by the film.

Explanation

Here ` W_0 = 0.6 eV `
For W_0 to be min ` ' lambda'` becomes maximum
` W_0 = hc/ lambda`
` or , `lambda = hc / W_0`
`= 6.63 xx 10^(-34) xx 3 xx 10^(8) / 0.6 xx 1.6 xx 10^(-19) `
` = 20.71 xx 10^(-7) m `
` = 2071 nm`