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The potential energy of a particle of mass ‘m’ is given by $U (x) = \{\begin{array}{l} E_{0} 0 \leq x \leq 1 \\ 0 x > 1 \end{array}\},$ $λ_{1}$ and $λ_{2}$ are the de-Broglie wave lengths of the particle, when 0 < x < 1 and x > 1 respectively. If the total mechanical energy of particle is $2 E_{0},$ find ${(\frac{λ_{1}}{λ_{2}})}^{2}$

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$\frac{1}{2}$

answer is B.

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Detailed Solution

Total energy=Potential energy+Kinetic energy

$P . E = U (x) = 0 \leq x \leq 1 \to 2 E_{0} = E_{0} + K . E_{1}$

$K . E_{1} = E_{0}$

$U (x) = 0; x > 1 \to 2 E_{0} = 0 + K E_{2}$

$K . E_{2} = 2 E_{0}$

$λ = \frac{h}{\sqrt{2 m (K . E)}}$

$λ \propto \frac{1}{\sqrt{K . E}} \Rightarrow λ^{2} \propto \frac{1}{K . E}$

${(\frac{λ_{1}}{λ_{2}})}^{2} = \frac{K . E_{2}}{K . E_{1}} = \frac{2 E_{0}}{E_{0}} = \frac{2}{1}$

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