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Photoelectric effect experiments are performed using three different metal plates p, and r having work functions phi_(P)=2.0eV,phi_(q)=2.5eV and wavelengths of 550 nm, 450 nm and 350 nm with equal intensities illuminates each of the plates. The correct I-V graph for the experiment is (Take hc = 1240 eV nm) |
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Answer»
Wavelength less than `lambda_("max")` alone will cause photelectrons to be ejected `lambda_(m)=(HC)/(phi)` For `phi_(p),lambda_(m_(p))=(1240eV nm)/(2.5eV)=496 nm` For `phi_(r),lambda_(m_(r))=(1240eV nm)/(3eV)=413.3 nm` Wavelngths in the incident beam are 550 nm, 450 nm and 350 nm. 350 nm waves can generate photoelectrons from p,q and r. 450 nm is shorter than `lambda_(m)` for p and q only and 550 nm `lt` 620 nm only in this group. so it can excite on p-cell. Currept `prop` intensity. Intensity =Nhv of photoelectrons. `:.` I is maximum for p cell, one gets the maximum intensity, and next is for q cell and the r cell can give photelectrons only by 350 nm. `:.` I is minimum for r. |
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