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The work function of tungsten is 4.52 e V. Find the cut-off potential for photoelectrons when a tungsten surface is illuminated with radiation of wavelength...

The work function of tungsten is 4.52 e V. Find the cut-off potential for photoelectrons when a tungsten surface is illuminated with radiation of wavelength 2.50 × 10-7 m. (Planck’s constant, h = 6.62 × 10-34 Js).

Answers

Mutiso
Frequency ‘f’ = c / ${\lambda}$ = 3.0 × 108 / 2.50 × 10-7.
Energy of photon = h f = 6.62 × 10-34 × (3.0 × 108 / 2.50 × 10-7) × (1 / 1.6 × 10-19) = 4.97 eV.
Hence hfco = 4.52 e V.
eVco = 4.97 e V - 4.52 e V = 0.45 e V = 7.2 × 10-20 J V co
= 7.2 × 10-20 / 1.6 × 10-19 = 0.45 e V.
Mutiso answered the question on October 17, 2018 at 05:49

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