When light of wavelength 120 nm falls on a gold surface, electrons having a maxi

Question

When light of wavelength 120 nm falls on a gold surface, electrons having a maximum kinetic energy of 5.25 eV are emitted. Find values for the following. the work function of gold Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. eV the cutoff wavelength Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error, nm the frequency corresponding to the cutoff wavelength Your response differs from the correct answer by more than 10%. Double check your calculations. Hz

Explanation / Answer

Work function is W = E-K_max = (h*c/lamda) - (5.25*1.6*10^-19) = ((6.625*10^-34*3*10^8)/(120*10^-9)) - (5.25*1.6*10^-19) = 8.1625*10^-19 = 5.101 eV

b) W = h*c/Lamda_0

cut off wavelength is lamda_o = h*c/w =(6.625*10^-34*3*10^8)/(8.1625*10^-19) = 243.5 nm

C) w = h*fo

fo = w/h = (8.1625*10^-19)/(6.625*10^-34) = 12.32E15 Hz

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