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The photoelectric work function of potassium bis 2.3eV. If light having a wavelength of 2800Å falls on potassium, find (a) The kinetic energy in electron volts of the most energetic electrons emitted. (b) The stopping potential in volts are |
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Answer» SOLUTION :Given, `W=2.3eV""lamda=2800Å` `:.E("in "EV)=(12400)/(lamda("in"Å))=(12400)/(2800)=4.4eV` `K_(MAX)=E-W=(4.4-2.3)eV=2.1eV` (b) `K_(max)=eV_(0):.2.1eV=eV_(0)(or)V_(0)=2.1` VOLT |
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