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If a photocell is illuminated with a radiation of `1240 A`, then stopping potential is found to be 8 V. The work function of the emitter and the threshold wavelength areA. 1 eV,`5200A`B. 2 eV,`6200A`C. 3 eV,`7200A`D. 4 eV,`4200A` |
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Answer» Correct Answer - B `W=hc-eV_S` `hupsilon=` energy of incident photon Here `hupsilon=(12400)/(1240)eV=10eV` `W=10-8=2eV` So, `lamda_0=` Threshold wavelength `=(12400)/(2eV)A=6200A` |
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