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Question
The work function of a photoelectric material is 4.0 eV. (a) What is the threshold wavelength? (b) Find the wavelength of light for which the stopping potential is 2.5 V.
(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
Work function of a photoelectric material, ϕ = 4 eV = 4 × 1.6 × 10−19 J
Stopping potential, V0 = 2.5 V
Planck's constant, `h = 6.63 xx 10^-34 "Js"`
(a) Work function of a photoelectric material,
`phi = (hc)/λ_0`
Here, λ0 = threshold wavelength of light
c = speed of light
`therefore λ_0 = (hc)/phi`
`λ_0 = (6.63 xx 10^-34 xx 3 xx 10^8)/(4 xx 1.6 xx 10^-19)`
`λ_0 = (6.63 xx 3)/64 xx (10^27)/(10^-9)`
`λ_0 = 3.1 xx 10^-7 "m"`
`λ_0 = 310 "nm"`
(b) From Einstein's photoelectric equation,
`E = phi + eV_0`
On substituting the respective values , we get :-
`(hc)/λ = 4 xx 1.6 xx 10^-19 + 1.6 xx 10^-19 xx 2.5`
`⇒ λ = (6.63 xx 10^-34 xx 3 xx 10^8)/(6.5 xx 1.6 xx 10^-19)`
`⇒ λ = (6.63 xx 3 xx 10^-26)/(1.6 xx 10^-19 xx 6.5)`
`⇒ λ = 1.9125 xx 10^-7 = 191 "nm"`
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