The potential energy of a particle of mass `m` is given by `V(x) = E_0` when `x= lex le 1 `and `xgt 1` repectively. `lambda_(1) and lambda_(2) ` are the de - Broglie wavelength of the particle, ,if the total energy of particle is `2 E_(0)` find `lambda_(1) // lambda_(2)`

The potential energy of a particle of mass `m` is given by `V(x) = E_0

1.	The potential energy of a particle of mass `m` is given by `V(x) = E_0` when `x= lex le 1 `and `xgt 1` repectively. `lambda_(1) and lambda_(2) ` are the de - Broglie wavelength of the particle, ,if the total energy of particle is `2 E_(0)` find `lambda_(1) // lambda_(2)`
Answer» Correct Answer - `sqrt(2)` For `0 lt x le1, " "KE=2E_(0)-E_(0)=E_(0)` for `x gt 1," " KE=2E_(0)" "(lambda_(1))/(lambda_(2))=(h//P_(1))/(h//P_(2))=(P_(2))/(P_(1))=sqrt((KE_(2))/(KE_(1)))=sqrt(2)`

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The potential energy of a particle of mass `m` is given by `V(x) = E_0` when `x= lex le 1 `and `xgt 1` repectively. `lambda_(1) and lambda_(2) ` are the de - Broglie wavelength of the particle, ,if the total energy of particle is `2 E_(0)` find `lambda_(1) // lambda_(2)`

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