Question

A photon with frequency 9.996x1014 Hz strikes a slab of lead and is absorbed without freeing an electron.

a) If the frequency of that photon is the threshold frequency for lead, calculate the work function for lead. Express your answer in eV.

b) Assume a photon of wavelength 200 nm strikes the lead, freeing an electron. Calculate the maximum kinetic energy (in Joules) of the liberated electron.

c) Calculate the de Broglie wavelength of the emitted electron. You will need to begin with the kinetic energy to determine the speed of the electron, which is not relativistic.

d) Assume the process described above occurs repeatedly, in rapid succession, and that the emitted electrons strike a double slit with a slit separation of 0.300 µm. The slits are 0.5 mm from the detection screen. Calculate the distance to the first and second interference maxima.

e) Calculate the first maximum (m=1) for light of wavelength 200 nm. How does this value compare to (m=1) for the electron?

Please answer all parts!

Answer 1

a. E = hf

Wo = E = 6.626*10^-34 * 9.996*10^14

Wo = 6.623*10^-19 J

Wo = 6.6263 *10^-19/(1.6*10^-19)

Wo = 4.14 eV  

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b. Energy associated with the wavelength 200 nm is

E = hc/L

E = 6.626*10^-34 * 3*10^8/(200*10^-9)

E = 9.93*10^-19 J

E = 6.211 eV

so Kmax = hf - Wo

Kmax = 6.211 - 4.14 = 2.071 eV

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c. wavelength associated with this electron is 0.5mv^2 =E = hc/L

L = h/sqrt(2m K)

L = (6.626*10^-34)/(sqrt(2* 9.11*10^-31* 2.071*1.6*10^-19)

L = 0.852 nm