1.

To change de-Broglie wavelength from 0.5xx10^(10)m to 1xx10^(-10)m its energy should be……

Answer»

<P>half of INITIAL KINETIC ENERGY
double of initial kineric energy
four times initial kinetic energy
`(1)/(4)` times initial kinetics energy

Solution :`lambda=(h)/(mv)=(h)/(p)`
but p =`sqrt(2mK)`
`therefore` In `lambda=(h)/(sqrt(2mK)),(h)/(sqrt(2m))` same
`therefore lambda prop (1)/(sqrt(K))`
`therefore K prop (1)/(lambda^(2))`
`therefore (K_(2))/(K_(1))=((lambda_(1))/(lambda_(2)))^(2)`
`=((0.5xx10^(-10))/(1xx10^(-10)))=((1)/(2))^(2)=(1)/(4)`
` therefore K_(2)=(K_(1))/(4)`


Discussion

No Comment Found

Related InterviewSolutions