If kinetic energy of particle is made 16 times,percentage change in its de-Broglie wavelength will be…….

Options

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

Answer

D

Explanation

de-Broglie wavelength ,
`lambda=(h)/(p)=(h)/(sqrt(2mK))`
`therefore lambda prop(1)/(sqrt(K))[because h,2,m` are same]
`therefore (lambda_(2))/(lambda_(1))=sqrt((K_(1))/(K_(2)))=sqrt((1)/(16))=(1)/(4)`
`therefore lambda_(2)=(lambda_(1))/(4)`
`therefore` Percentage change,
`=(lambda_(2)-lambda_(1))/(lambda_(1))xx100=((lambda_(1))/(4)-lambda_(1))/(lambda_(1))xx100%`
`=-(3lambda_(1))/(4lambda_(1))xx100%` `therefore` wavelength decrease by 75%