InterviewSolution
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InterviewSolution
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A stream of electrons accelerated by a potential difference `V` falls on the surface of a metal whose inner potential is `V_(i)= 15V`. Find: (a) the refractive index of the metal for the electrons accelerated by a potential difference `V= 150V`, (b) the values of the raito `V//V_(i)` at which the refracticve index differs from unity by not more than `eta=1.0%`. |
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Answer» Inside the metal, there is a negative potential energy of `-eV_(i)`. (This potential energy prevents electron from leaking out and can be measured in photoelectric effect ets.)An electron whose `K.E` is `eV` outside the metal will find its `K.E` increased to `e(V+V_(i))` in the metal. Then (a) de Broglie wavelength in the metal `=lambda_(m)=(2piħ)/(sqrt(2m e (V+V_(i))))` Also de Broglie wavelength in Vacuum `=lambda_(0)=(2pi ħ)/(sqrt(2 mVe))` Hence refractive index `n=(lambda_(0))/(lambda_(m))=sqrt(1+(V_(i))/(V))` Substituting we get `n=sqrt(1+(1)/(10))~=1.05` (b) `n-1= sqrt(1+(V_(i))/(V))-1leeta` then `1+(V_(i))/(V)le(1+eta)^(2)` or `V_(i)leeta(2+eta)V` or `(V)/(V_(i))ge(1)/(eta(2+eta))` For `eta= 1% = 0.01` we get `(V)/(V_(i))ge50` |
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