Question #76750
a photon of red light wavelength 470 nm falls on a metal surface. as a result photoelectrons are ejected with a velocity of 6.4×x10^4 m/s . find (a) kinetic energy of emitted photoelectrons.(b)the work function of the metal surface
Expert's answer
Wavelength of electron = 470 nm
velocity of electron= 6.4×x10^4 m/s
Kinetic energy of electron is
= 1/2mv^2
=1/2 9.1×10^-31×6.4×10^4×6.4×10^4
=.00186×10^-18 J
Work function of electron
W=hc/wavelength -1/2mv2
W=6.626×10^-34 ×3×10^8/470×10^-9 - .00186 × 10-18 J
W= .42104×10^-18 J
velocity of electron= 6.4×x10^4 m/s
Kinetic energy of electron is
= 1/2mv^2
=1/2 9.1×10^-31×6.4×10^4×6.4×10^4
=.00186×10^-18 J
Work function of electron
W=hc/wavelength -1/2mv2
W=6.626×10^-34 ×3×10^8/470×10^-9 - .00186 × 10-18 J
W= .42104×10^-18 J
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