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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. `2`B. `1`C. `sqrt(2)`D. `(1)/(sqrt(2))` |
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Answer» Correct Answer - C `K.E. = 2 E_(0) - E_(0) = E_(0) (for 0le x le 1)` `rArr lambda_(1) = (h)/(sqrt( 2 m E_(0))` `K.E. = 2 E_(0) ( for x gt 1) rArr lambda_(2) = (h)/(sqrt( 4 m E_(0))) rArr (lambda_(1))/(lambda_(2)) = sqrt(2)` |
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