The potential energy of a particle of mass `m` is given by `V(x) = lambda_(1) and lambda_(2) ` are the de - Brogle wavelength of the particle, when `= lex le 1 `and `xgt 1` repectively ,if the total energy of particle is `2 E_(0) find lambda_(1) // lambda_(2)`
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The de brogle wave length is given by `lambda = (h)/(m v) rArr lambda = (h)/(sqrt(2mK))` cose (i) `0 le xx le 1` For ,this potential energy is `E_(0)` (given) Total energy = 2E_(0)`(given) `:. kinetic energy `= 2E_(0) - E_(0) = E_(0) ` `lambda_(1) = (h)/(sqrt(2mE_(0))) `...(ii) Dividing (i) and (ii) `(lambda_(1))/(lambda_(2)) = sqrt((2 E_(0))/(E_(0))) rArr (lambda_(1))/(lambda_(2)) = sqrt(2)`
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