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If the kinetic energy of the particle is increased to `16` times its previous value , the percentage change in the de - Broglie wavelength of the particle isA. `25`B. `75`C. `60`D. `50` |
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Answer» Correct Answer - B As we know `lambda = (h)/(p) = (h)/(sqrt(2 mK))` `(because P = sqrt( 2mKE))` or `(lambda_(1))/(lambda_(2)) = sqrt((K_(2))/(K_(1))) = sqrt((16 K)/( K)) = (4)/(1)` Therefore the percentage change in de - Broglie wavelength ` = ( 1 - 4)/(4) xx 100 = - 75%` |
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