1.

If the kinetic energy of a particle is increased by 16 times, the percentage change in the de-Broglie wavelength of the particle is

Answer»

25
75
60
50

Solution :`lambda = (h)/(SQRT(2mK)) , lambda. = (h)/(sqrt(2m XX 16 K)) = (lambda)/(4)`
% change in de-Broglie wavelength, `(lambda - lambda.)/(lambda) = (1 - (lambda.)/(lambda)) xx 100 = (1-(1)/(4)) xx 100 = 75%`


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