Physics, asked by patgars, 11 months ago

A photon of wavelength 99 nm falls on a metal plate of work function 1.0 eV. The wavelength of the fastest emitted photoelectron is (given h = 6.6 ×10–34 Js, mass of electron (m = 9.1×10–31 kg)

Answers

Answered by KomalSrinivas
0

The wavelength of the fastest emitted photoelectron is  3.6A°

Wavelength of the photons = 99nm

                                           = 99 * 10 ('-9) m

Energy of the photon = (hc)/Д ( Д =Wavelength)

                                     = 2 * 10('-11) J

Ke max = hv-Work function

           Work function = 1 ev

Ke max = 2 * 10('-18) - 1.6 * 10('-19)

            = 1.84 * 10('-18)

Ke max = p²/2m (p = maximum momentum)

From de broglie Д = h/p

p = √(2m kEmax)

Д = 3.6 A°

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