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The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is themaximum kinetic energy of photoelectrons emitted? |
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Answer» SOLUTION :`V_(0)=1.5V` `K_("max")=(1)/(2)mv_("max")^(2)=eV_(0)=1.6xx10^(-19)xx1.5=2.4xx10^(-19)J` |
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