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| 1. |
The photoelectric cut-off voltage in certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted ? |
| Answer» Give, `V_(0)=1.5 V, e = 1.6xx10^(-19)C, KE_("max")=eV_(0)=1.6xx10^(-19)xx1.5=2.4xx10^(-19)J`. | |