Question

The potential energy of particle of mass m varies as
`U(x)={(E_(0)"for"0lexle1),(0" for " gt 1):}`
The de Broglie wavelength of the particle in the range `0lexle1 " is " lamda_(1)` and that in the range `xgt1" is "lamda_(2)`.
If the total of the particle is `2E_(0)," find "lamda_(1)//lamda_(2)`.
A. `sqrt(2)`
B. `sqrt(3)`
C. `sqrt((1)/(2))`
D. `sqrt((2)/(3))`

Answer 1

Correct Answer - A
(a) : de Broglie wavelength of particle of mass m is
`lamda=(h)/(sqrt(2mK))` where K is the kinetic energy of the particle.
`0lexle1,U(x)=E_(0)`
As total energy = Kinetic energy + Potential energy
`:.2E_(0)=K_(1)+E_(0)""( :."Total energy"=2E_(0)"(Given))"`
or `lamda_(1)=(h)/(sqrt(2mK_(1)))=(h)/(sqrt(2mE_(0)))` . . . (i)
For `xgt1,U(x)=0" ":." "K_(2)=2E_(0)`
`:.lamda_(2)=(h)/(sqrt(2mK_(2)))=(h)/(sqrt(2E_(0)))` . . . (ii)
Divide (i) by (ii), we get
`(lamda_(1))/(lamda^(2))=(h)/(sqrt(2mE_(0)))xx(sqrt(2m(2E_(0))))/(h)=sqrt(2)`