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.

Explanation

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`