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Question
Monochromatic radiation of wavelength 640.2 nm (1 nm = 10−9 m) from a neon lamp irradiates photosensitive material made of caesium on tungsten. The stopping voltage is measured to be 0.54 V. The source is replaced by an iron source and its 427.2 nm line irradiates the same photo-cell. Predict the new stopping voltage.
Solution
Wavelength of the monochromatic radiation, λ = 640.2 nm = 640.2 × 10−9 m
Stopping potential of the neon lamp, V0 = 0.54 V
Charge on an electron, e = 1.6 × 10−19 C
Planck’s constant, h = 6.6 × 10−34 Js
Let `phi_0` be the work function and ν be the frequency of emitted light.
We have the photo-energy relation from the photoelectric effect as:
eV0 = hv − `phi_0`
`phi_0 = "hc"/lambda - "eV"_0`
= `(6.6 xx 10^(-34) xx 3 xx 10^8)/(640.2 xx 10^(-9)) - 1.6 xx 10^(-19) xx 0.54`
= `3.093 xx 10^(-19) - 0.864 xx 10^(-19)`
= `2.229 xx 10^(-19) "J"`
`= (2.229 xx 10^(-19))/(1.6 xx 10^(-19))`
= 1.39 eV
Wavelength of the radiation emitted from an iron source, λ' = 427.2 nm
= 427.2 × 10−9 m
Let `"V"_0^"'"` be the new stopping potential. Hence, photo-energy is given as:
`"eV"_0^"'" = "hc"/(lambda"'") - phi_0`
= `(6.6 xx 10^(-34) xx 3 xx 10^(8))/(427.2 xx 10^(-9)) - 2.229 xx 10^(-19)`
= `4.63 xx 10^(-19) - 2.229 xx 10^(-19)`
= `2.401 xx 10^(-19) "J"`
`= (2.401 xx 10^(-19))/(1.6 xx 10^(-19))`
= 1.5 eV
Hence, the new stopping potential is 1.50 eV.
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