The potential energy of a particle of mass m is given by λ1 and λ2 are the de-Broglie wavelengths of the particle, when 0 x 1 and x > 1 respectively. If the total energy of particle is 2E0 ,the ratio will be

K.E.= 2 E0– E0 = E0 (for 0 x 1) ⇒ K.E. = 2 E0 (for x > 1) ⇒

The potential energy of a particle of mass m is given by λ1 and λ2 are the de-Broglie wavelengths of the particle, when 0 x 1 and x > 1 respectively. If the total energy of particle is 2E0 ,the ratio will be

K.E.= 2 E0– E0 = E0 (for 0 x 1) ⇒ K.E. = 2 E0 (for x > 1) ⇒

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The potential energy of a particle of mass m is given by
$[U(x) = left{ {begin{array}{{20}{c}} {{E_0};} \ {0;} end{array},,,begin{array}{{20}{c}} {0 leqslant x leqslant 1} \ {x > 1} end{array}} right.,,,,,,$
λ₁and λ₂ are the de-Broglie wavelengths of the particle, when 0 $leq$ x $leq$ 1 and x > 1 respectively. If the total energy of particle is 2E₀ ,the ratio $frac{{{lambda _1}}}{{{lambda _2}}}$ will be

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$sqrt 2$

$frac{1}{{sqrt 2 }}$

answer is C.

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K.E.= 2 E₀– E₀ = E₀ (for 0 $leq$ x $leq$ 1) ⇒ ${lambda _1} = frac{h}{{sqrt {2m,{E_0}} }}$
K.E. = 2 E₀(for x > 1) ⇒ ${lambda _2} = frac{h}{{sqrt {4m,{E_0}} }} Rightarrow frac{{{lambda _1}}}{{{lambda _2}}} = sqrt 2 .$

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The potential energy of a particle of mass m is given by λ1 and λ2 are the de-Broglie wavelengths of the particle, when 0 x 1 and x > 1 respectively. If the total energy of particle is 2E0 ,the ratio will be

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