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)`A. `sqrt2`B. `1/sqrt2`C. `2:1`D. `4:1`

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)`A. `sqrt2`B. `1/sqrt2`C. `2:1`D. `4:1`
Answer» Correct Answer - A `lambda_(1)/lambda_(2)=(h)/(p_(1)/(h))=p_(2)/p_(1)=sqrt(2mk_(2))/(sqrt(2mk_(1)))=sqrt(2E_(0))/sqrt(2E_(0)-E_(0))=sqrt2`

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