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InterviewSolution
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Red light, howsoever bright it is, cannto produce the emission of electrons from a clean zinc surface, but even weak ultraviolet radiation can do so. Why? X-rays of wavelength lamda fall on photosensitive surface, emitting electrons. Assuming that the work function of the surface can be neglected, prove that the de-Broglie wavelength of electrons emitted will be sqrt((hlamda)/(2mc)). |
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Answer» Solution :For photoelectric emission, the energy of incident photon MUST be greater than the work function of the METAL. For zinc work function is about 3.4 eV. As energy of red light photon is less hence photo-emission will not TAKE PLACE. However, having much smaller wavelength the energy of ultraviolet light photon is much higher and, consequently, it can cause emission of electrons, even if its intensity is weak. If we neglect work function of the photosensitive surface, then kinetic energy of photo-electrons will be `K=hv=(hc)/(LAMDA)` `therefore`de-Broglie wavelength of these photoelectrons will be `lamda_("electron")=(h)/(sqrt(2mK))=(h)/(sqrt(2m*(hc)/(lamda)))=sqrt((hlamda)/(2mc))`. |
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