If photoelectrons are to be emitted from a potassium surface with a speed `6 xx 10^(6) ms^(-1)`, what frequency of radiation must be used ? (Threshold frequency for potassium is `4.22 xx 10^(14)` Hz, h = `6.6 xx 10^(-34)` Js, `m_(e) = 9.1 xx 10^(-31)` kg)

Explanation

Here, `v = 6 xx 10^(6) ms^(-1)`
`V_(0) = 4.22 xx 10^(14)` Hz
From Einstein.s photoelectric equation,
`K.E = (1)/(2) mv^(2) = h(upsilon - upsilon_(0))`
`upsilon = (1)/(2) (mv^(2))/(h) + upsilon_(0)`
`= (1)/(2) xx (9.1 xx 10^(-31) + (6 xx 10^(6))^(2))/(6.6 xx 10^(-34)) + 4.22 xx 10^(14)`
`= (2.48 xx 10^(14)) + (4.22 xx 10^(14))`
`upsilon = 6.7 xx 10^(14)` Hz