Saved Bookmarks
| 1. |
If work function for certain metal is 1.8ev what is stopping potential for electrons ejected from metal when light of 4000angstrom minute shine on metal ? what is speed of electron |
|
Answer» V= 1.3VExplanation:We know the RELATION hf = K max + Ø where photon energy is h F, k max is kinetic energy of the emitted electrons and WORK function is Ø. 4000 Angstrom = 400 nanometreSo K max = e VNow h f = e V + ØSo frequency of incident LIGHT is f = C / λh c / λ = e V + ØTo solve for stopping potential we get h c / λ – eV = Ø – e V = Ø – h c / λV = Ø / e – h c / e λ V = h c / e λ – Ø / eSubstituting the value we getV = 6.62 x 10^-34 x 2.99 x 10^8 / 1.602 x 10^-10 x 400 x 10^-9 - 1.8 V x e / e3.1 V – 1.8 VV = 1.3 V |
|