When a certain metal was irradiated with a light of frequency 3.2 xx 10^(16) Hz, the photoelectrons had twice the kinetic energy as emitted when the same was irradiated with light of frequency 2.0 xx 10^(16)Hz. Calculate the threshold frequency (v_(0)) of the metal.

When a certain metal was irradiated with a light of frequency 3.2 xx 1

1.	When a certain metal was irradiated with a light of frequency 3.2 xx 10^(16) Hz, the photoelectrons had twice the kinetic energy as emitted when the same was irradiated with light of frequency 2.0 xx 10^(16)Hz. Calculate the threshold frequency (v_(0)) of the metal.
Answer» Solution :Kinetic energy of photoelectrons emitted `= hv - hv_(0) = h (v - v_(0))` In 1ST case, `(K.E.)_(1) = h (3.2 XX 10^(16) - v_(0))` In 2ND case, `(K.E.)_(2) = h (2.0 xx 10^(16) - v_(0))` But `(K.E.)_(1) = 2 (K.E.)_(2) ("Given") :. h (3.2 xx 10^(16) - v_(0)) = 2h (2.0 xx 10^(16) - v_(0))` or `v_(0) = 4 xx 10^(16) - 3.2 xx 10^(16) = 0.8 xx 10^(16) = 8 xx 10^(15) HZ`