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The potential energy of a partical varies as .`U(x) = E_0 ` for ` 0 le x le 1``= 0` for `x gt 1 ` for `0 le x le 1` de- Broglie wavelength is `lambda_1` and for `xgt1` the de-Broglie wavelength is `lambda_2`. Total energy of the partical is `2E_0`. find `(lambda_1)/(lambda_2).`A. `sqrt(2)`B. `(1)/(sqrt(2))`C. 2D. `(1)/(2)` |
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Answer» Correct Answer - A For `0 le x le 1, KE = 2E_(0) - E_(0) =E_(0)` for `x gt 1` `(lamda_(1))/(lamda_(2))=(h//P_(1))/(h//P_(2))` `=(P_(2))/(P_(1))=sqrt((KE_(2))/(KE_(1)))` `= sqrt((2E_(0))/(E_(0)))=sqrt(2)` |
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