1. Molybdenum has a work function of 4.2 eV. Find the cutoff wavelength for the photoelectric effect. The speed of light is 3 × 108 m/s and Planck’s constant is 6.626 × 10−34 J · s. Answer in units of nm.
2. Calculate the stopping potential if the incident light has a wavelength of 109 nm. Answer in units of V.
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wavelength, λ = 1240/4.2 = 295 nm
frequency = c / λ= (3x10^8) / (295x10^-9) = 1.0169x10^15 Hz
wavelength (λ ) of the incident light = 109 nm therefore
Energy E = hc/λ = [4.135x10^(-15)* 3x10^8 ] /(109x10^(-9) = 11.38 eV
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