The work function for potassium is 2.24 eV. (a) If potassium metal is illuminate
ID: 1757915 • Letter: T
Question
The work function for potassium is 2.24 eV. (a) If potassium metal is illuminated with light of wavelength320 nm, find the maximum kinetic energy of thephotoelectrons. The speed of light is 3 x 108 m/sand Planck's constant is 6.62607 x 10-34 J s.Answer in units of eV. (b) Find the cutoff wavelength. Answer in units of nm. (a) If potassium metal is illuminated with light of wavelength320 nm, find the maximum kinetic energy of thephotoelectrons. The speed of light is 3 x 108 m/sand Planck's constant is 6.62607 x 10-34 J s.Answer in units of eV. (b) Find the cutoff wavelength. Answer in units of nm. (b) Find the cutoff wavelength. Answer in units of nm.Explanation / Answer
The work function for potassium W = 2.24 eV. (a) wavelength = 320 nm = 320 * 10 ^ -9 m The speed of light c = 3 x108 m/s Planck's constant h = 6.62607 x10-34 J s Maximum kinetic energy = Energy of incident photon - workfunction ------( 1) Energy of incedent photon = hc / = 62.119 * 10 ^ -20 J = 62.119 * 10 ^ -20 / ( 1.6 * 10 ^ -19 ) eV = 3.8824 eV from eq ( 1 ) , maximum kinetic energy = 3.8824 eV -2.24 eV = 1.6424 eV (b). cutoff wavelength ' = hc / W = [ ( 6.62607 * 10 ^ -34 Js ) * ( 3 * 10 ^ 8 m / s ) ] / [ 2.24 *1.6 * 10 ^ -19 J ] = 554.637 * 10 ^ -9 m = 554.637 nm (a) wavelength = 320 nm = 320 * 10 ^ -9 m The speed of light c = 3 x108 m/s Planck's constant h = 6.62607 x10-34 J s Maximum kinetic energy = Energy of incident photon - workfunction ------( 1) Energy of incedent photon = hc / = 62.119 * 10 ^ -20 J = 62.119 * 10 ^ -20 / ( 1.6 * 10 ^ -19 ) eV = 3.8824 eV from eq ( 1 ) , maximum kinetic energy = 3.8824 eV -2.24 eV = 1.6424 eV (b). cutoff wavelength ' = hc / W = [ ( 6.62607 * 10 ^ -34 Js ) * ( 3 * 10 ^ 8 m / s ) ] / [ 2.24 *1.6 * 10 ^ -19 J ] = 554.637 * 10 ^ -9 m = 554.637 nmRelated Questions
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