Question

The threshold frequency, `v_(0)`, of a metal is `6.7xx10^(14)s^(-1)`. Calculate the maximum kinetic enegry of a single electron that is emitted when a radiation of frequency `v = 1.0 xx 10^(-15)s^(-1)` strikes the metal.
Strategy: Use the relatiship between `v_(0),v,`and `KE` given in Eq.

Answer 1

Step `I.` The relatiship between `v_(0),v`, aand maximum kinetic enegry is given as
`KE_(max) = hv - hv_(0)`
Step `2`. Solving the equation for kinetic energy gives
`KE_(max) = h(v - v_(0))`
`=(6.6xx10^(-34)Js)(1.0 xx 10^(15)s^(-1)-6.7xx10^(14)s^(-1))`
`=(6.6xx10^(-34) Js) (10xx10^(14)s^(-1) - 6.7 xx 10^(14) s^(-1))`
`=(6.6 xx 10^(-34)Js) (3.3xx10^(14)s^(-1))`
`= 2.2xx10^(-19)J`