Photoelectric emission is observed from a metal surface with incident frequencies v_(1) and v_(2) where v_(1) gt v_(2). If the kinetic energies of the photoelectrons emitted in the two cases are in the ratio 2 : 1, then the threshold frequency v_(0) of the metal is

Photoelectric emission is observed from a metal surface with incident

1.	Photoelectric emission is observed from a metal surface with incident frequencies v_(1) and v_(2) where v_(1) gt v_(2). If the kinetic energies of the photoelectrons emitted in the two cases are in the ratio 2 : 1, then the threshold frequency v_(0) of the metal is
Answer» `v_(1) = v_(2)` `(v_(1) - v_(2))/(H)` `2 v_(1) - v_(2)` `2v_(2) - v_(1)` Solution :`hv_(1) = hv_(0) + E` (K.E)..(i) `hv_(2) = hv_(0) + (1)/(2) E or 2 hv_(2) = 2 hv_(0) + E`...(ii) EQN. (i) - Eqn (ii) gives `hv_(1) - 2hv_(2) = hv_(0) - 2 hv_(0)` or `v_(1) - 2 v_(2) = - v_(0) or v_(0) = 2 v_(2) - v_(1)`