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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. 75B. 60C. 50D. 25 |
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Answer» Correct Answer - A de-Broglie wavelength `lambda =(h)/(p) =(h)/(sqrt(2mK))` `(lambda_(2))/(lambda_(1)) =sqrt((K_(1))/(K_(2))) =sqrt((K)/(16K)) =(1)/(4)` `%` changed `=((lambda_(2))/(lambda_(1))-1)xx100= ((1)/(4)-1)xx100=-75%` i.e. `75%` decrease |
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