InterviewSolution
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InterviewSolution
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In accordance with classical electrodynamic an electron moving with acceleration `W` loses its energy due to radiation as `(dE)/(dt)= -(2e^(2))/(3c^(3))W^(2)`, Where `e` is the electron, `c` is the velocity of light. Estimate the time during which the energy of an electron performing alomist harmonic oscillations with frequency `omega= 5.10^(15)s^(-1)` will decrease `eta= 10 time s` |
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Answer» The formula in `MKS` units is `(dE)/(dt)= (mu_(0)e^(2))/(6pic)w^(rarr_2)` For an electron performing (linear) harmonic vibrations `vec(w)` is the some definite direction with `w_(x)=-w^(2)x` say. Thus `(dE)/(dt)=-(mu_(0)e^(2)omega^(4))/(6pic)x^(2)` If the radiation loss is small (i.e., if `omega` is no to large), then the motion of the electron is always close to simple harmonic wiht slowly decreasing amplitude. Then we can write `E=(1)/(2)m omega^(2)a^(2)` and `x=a cos omegat` and avarage the above equation ignoring the variation of `a` in any cycle. Thus we get the equation, on using `lt x^(2) ge(1)/(2)a^(2)` `(dE)/(dt)=-(mu_(0)e^(2)omega^(4))/(6pic).(1)/(2)a^(2)=-(mu_(0)e^(2)omega^(2))/(6pi mc)E` since `E=(1)/(2)momega^(2)a^(2)` for a harmonic oscillator. This equation intergates to `E=E_(o)e^(-t//T)` where `T= 6pi mc//e^(2)omega^(2)mu_(0)` It is than see that energy decreases `eta` times in `t_(0)=T In eta=(6pi mc)/(e^(2)omega^(2)mu_(0))In eta= 14.7ns` |
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