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The additional energy that should be given to an electron to reduce its de-Broglie wavelength from 1 mm to 0.5 mm is: |
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Answer» 2 times the initial kinetic energy `lambda =H/sqrt(2m E_(x)) [therefore lambda=(1)/sqrtE_(k)]` (h and m remain constant), Here, in given condition `lambda_(1)/lambda_(2) =sqrt(K_(k_(2))/(E_(k_(1))), lambda_1/lambda_(2)=sqrt((4E)/(E)))` `lambda_(1)=2lambda_(2), lambda_(2)=lambda_(1)/2` So, if `E_(k)` is increased by four times, then `lambda` becomes half. Additional kinetic energy that should be SUPPLIED to the ELECTRON. `=4E_(k)-E_(k)=2E_(k)` |
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