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

(b) : As `lamda=(h)/(mv)=(h)/(sqrt(2mK))orlamdaprop(1)/(sqrt(K))`
`:.(lamda^('))/(lamda)=sqrt((K)/(K^(')))=sqrt((1)/(16))=(1)/(2)`
`%` change in de Broglie wavelength `=((lamda-lamda^('))/(lamda))xx100=(1-(lamda^('))/(lamda))100=(1-(1)/(4))xx100=75%`