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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» 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` |
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