Question Number 61993 by necx1 last updated on 13/Jun/19
$${ultraviolet}\:{light}\:{of}\:{wavelength}\:\mathrm{300}×\mathrm{10}^{−\mathrm{9}} {m} \\ $$$${causes}\:{photon}\:{emissions}\:{from}\:{a}\:{surface} \\ $$$${The}\:{stopping}\:{potential}\:{is}\:\mathrm{6}{V}.{Find}\:{the} \\ $$$${work}-{function}\:{in}\:{electron}-{Volts} \\ $$
Answered by ajfour last updated on 13/Jun/19
$${h}\nu=\phi+{eV}_{\mathrm{0}} \\ $$$$\phi=\frac{{hc}}{\lambda}−{eV}_{\mathrm{0}} \\ $$$$\:\:\:=\:\left(\frac{\mathrm{6}.\mathrm{626}×\mathrm{10}^{−\mathrm{34}} ×\mathrm{3}×\mathrm{10}^{\mathrm{8}} }{\mathrm{1}.\mathrm{6}×\mathrm{10}^{−\mathrm{19}} ×\mathrm{300}×\mathrm{10}^{−\mathrm{9}} }−\mathrm{6}\right){eV} \\ $$$$\:\:\:=\:\mathrm{4}.\mathrm{14}−\mathrm{6}\:<\mathrm{0} \\ $$$$\left({Wrong}\:{data}!\right). \\ $$