PhysicsEasy145×since 2002Q6622Radiation of wavelength λ,\lambda ,λ, is incident on a photocell. The fastest emitted electron has speed v.v.v. If the wavelength is changed to 3λ4,{{3\lambda } \over 4},43λ, the speed of the fastest emitted electron will be:A=v(43)12= v{\left( {{4 \over 3}} \right)^{{1 \over 2}}}=v(34)21B=v(34)12= v{\left( {{3 \over 4}} \right)^{{1 \over 2}}}=v(43)21C>v(43)12> v{\left( {{4 \over 3}} \right)^{{1 \over 2}}}>v(34)21D<v(43)12< v{\left( {{4 \over 3}} \right)^{{1 \over 2}}}<v(34)21Check answerSkip