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In question number 5, find the kinetic energy of the most energetic photoelectron emitted at t = 10 s when it reaches plate B. (Neglect the time taken by the photoelectron to reach plate B) |
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Answer» 23 eV `=8xx10^(-12)` Charge on B, `Q_(B)=(33.7-8)xx10^(-12)C=25.7xx10^(-12)C` `:.E=(sigma_(B))/(2epsilon_(0))-(sigma_(A))/(2epsilon_(0))ORE=(1)/(2Aepsilon_(0))(Q_(B)-Q_(A))` or `E=(1.77xx10^(-12))/(2xx(5xx10^(-4))xx(8.85xx10^(-12)))` or `E=2000N//C` Energy of photoelectrons on plate B energy = E-W=(5-2)eV=3eV Increase in energy `=(Ed)eV=(2xx10^(3))(10^(-2))eV=20eV` `:."Energy of photoelectrons on plate"` B=(20+3)eV=23eV |
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