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)`.