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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.
(Use h = 6.63 × 10-34J-s = 4.14 × 10-15 eV-s, c = 3 × 108 m/s and me = 9.1 × 10-31kg)
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Solution
Given:-
Work function, `W_0` = 0.6 eV
Now, work function,
`W_0 = (hc)/λ`,
where, h = Planck's constant
λ = wavelength of light
c = speed of light
`therefore λ = (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"`
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