The work function for potassium is 2.24 eV. If potassiummetal is illuminated wit

Question

The work function for potassium is 2.24 eV. If potassiummetal is illuminated with light of wavelength 350 nm, find themaximum kinetic energy of the photoelectrons. The speedof light is 3 × 108 m/s and Planck’s constant is6.62607 × 1034 J · s . Answer in units ofeV.
Find the cutoff wavelength. Answer in units of nm.
The work function for potassium is 2.24 eV. If potassiummetal is illuminated with light of wavelength 350 nm, find themaximum kinetic energy of the photoelectrons. The speedof light is 3 × 108 m/s and Planck’s constant is6.62607 × 1034 J · s . Answer in units ofeV.
Find the cutoff wavelength. Answer in units of nm.

Explanation / Answer

given W 0= 2.24ev = 35*10-9m maxium kinetic energy = K.Emax = hc/-W0    6.62607 × 1034 J · s.3*108/350*10-9 - 2.24ev solve K.Emax = 0.056*10-17 - 2.24*1.6*10-19 solve = -------ev cuttof wave length = hc/W0 plug in values 0 = -----nm plug in values 0 = -----nm

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