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In an experiment on photo electric emission, following observations were made, (i) Wavelength of the incident light `=1.98xx10^(-7)m`, (ii) Stopping potential `=2.5 "volt"`. Find: (a) Kinetic energy of photoelectrons with maximum speed. (b) Work function and (c )Threshold frequency, |
Answer» (a) Since `V_(s)=2.5V`, `K_(max)=eV_(s)` so, `K_(max)=2.5 eV` (b) Energy of incident photon `E=(12400)/(1980)eV=6.26 eV` , `W=E-K_(max)=3.76 eV` (c ) `hv_(th)=W=3.76xx1.6xx10^(-19)J :. v_(th)=(3.76xx1.6xx10^(-19))/(6.6xx10^(-34))~~9.1xx10^(14)Hz` |
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