InterviewSolution
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InterviewSolution
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A meacury arec lamp provides `0.1` watt of ultra-violet radiation at a wavelength of `lambda=2537 Å` only. The photo tube (cathode of photo electric device) consists of potessium and has an effective area of `4 cm^(2)`. The cathode is located at a distance of `1 m` from the radiation source. The work funcation for potassium is `phi_(0) = 2.22 eV`. According to classical theory, the radiation from arc lamp spreads out uniformaly in space as spherical wave. What time of expource to the radiation should be required for a potassium atom (radius `2Å`) ub tge cathode to accumulate sufficient enerfy to eject a photo-electron?A. `352` secondB. `176` secondC. `704` secondsD. None time lag |
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Answer» Correct Answer - A UV energy flux at a distance of `1m = (0.1)/(4pi xx 1^(2))` cross section (effective area) of atom `= pi xx (2 xx 10^-10)^(2) = 4pi xx 10^(20) m^(2)` Energy required to eject a photoelectron from potassium `= 2.2 eV -= 2.2 xx 1.6 xx 10^(-19) J`. `rArr "Exposure time" = (2.2 xx 1.6 xx 10^(-19))/(((0.1)/(4pi xx 1^(2))) (4pi xx 10^(-20))) = 352` seconds |
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