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 is

Options

A. 25
B. 75
C. 60
D. 50

Answer

B

Explanation

de Broglie wavelength , `lamda=h/(2mK)`
where m is te mass and K is the kinetic energy of the particle, When kinetic energy of the praticle is increased to 16 times, then its de Broglie wavelength becomes ,
`lamda. = h/(sqrt(2m(16K)))=1/4 lamda/(sqrt(2mK))=lamda/4" "` (Using (i))
% change in the de Broglie wavelength
`=(lamda-lamda.)/lamdaxx100 =(1-(lamda.)/lamda)xx100 =(1-1/4)xx100 =75% `