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(a) Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of 500 V with respect to the emitter. Ignore the small initial speeds of the electrons. The specific charge of the electron i.e., its e/m is given to be 1.76xx10^(11)C" "kg^(-1). (b) Use the same formula you employ in (a) to obtain electron speed for an collector potential of 10 MV. do you see what is wrong ? In what way is the formula to be modified ? |
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Answer» Solution :(a) Here accelerating voltage `V=500V and (e)/(m)` for electrons`=1.76xx10^(11)C " "kg^(-1)` As `eV=(1)/(2)mv^(2)impliesv=sqrt((2eV)/(m))` `therefore v=sqrt(2XX(1.76xx10^(11))xx(500))=1.33xx10^(7)MS^(-1)` (b) If accelerating voltage `V=10MV=10 xx10^(6)V+10^(7)V`, then using same formula as in (a), we GET `v=sqrt(2xx(1.76xx10^(11))xx10^(7))=1.88xx10^(9)ms^(-1)` However, the answerr obtained is wrong because no material particle can ever travel with a speed greater than the speed of light having a value `c=3xx10^(8)ms^(-1)`. The formula for kinetic energy `K=(1)/(2)mv^(2)` is, in fact, valid only when `(v)/(c) LT lt 1.` At high speeds when `v/c` is comparable to 1 (although, it is still below 1), we should apply Einstein.s relativic formula. On applying Einstein.s relativistic formulae, it is found that for 10 MeV energy electron the speed is 99.9% of the speed of light. |
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