InterviewSolution
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InterviewSolution
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X-rays of wavelength `lambda` fall on photosenstive 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((hlambda)/(2mc))`. |
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Answer» As work function of a surface can be neglected so `phi_0=0`. Therefore, kinetic energy of emitted photoelectron `1/2mv^2=(hc)/lambda-phi_0=(hc)/lambda or mv=sqrt((2mhc)/lambda)` de-Broglie Wavelength of emitted photoelectron `lambda_1=h/(mv)=h/(sqrt((2mhc)//lambda))=sqrt((hlambda)/(2mc))` proved |
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