Stopping potential question?

Question

The stopping potential in a photoelectric experiment is 2.3V when the illuminating radiation has wavelength 365nm. What is the work function of the emitting surface?

Answer 1

photon energy = overcome surface barrier + KE of electrons

hc/ λ = phi (work function) + 0.5 m v^2

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stopping potential (Vo) > when applied will just make the speed of ejected electron to zero on reaching the plate

so >0.5 mv^2 = KE of electrons = e Vo

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hc/ λ = phi (work function) + e Vo

phi = hc/ λ - eVo

= [6.6*10^-34 * 3*10^8/365*10^-9] - (1.6*10^-19*2.3)

= [5.43*10^-19] - (3.68*10^-19)

phi = 1.75*10^-19 Joule

phi = 1.75*10^-19 /1.6*10^-19 = 1.09 electron-volts

Answer 2

Stopping Potential

Answer 3

KE of electron = hf - phi = qV



so phi = hc / lambda - 2.3 eV.



They give you the wavelength (lambda). Look up planck's constant (prefereable in eV-s to avoid conversion pain), look up c. Plugnchug.