The potential energy of a particle of mass ‘m’ varies with its position ‘x’ as The de-Broglie wavelengths of the particle when 0≤x≤1 and x>1 are λ1 and λ2 respectively. If the total energy of the particle is 2E0, then the value of λ1λ2 is

K.E. =2E0−E0=E0( for 0≤x≤1)⇒λ1=h2mE0 K.E. =2E0( for x>1)⇒ λ2=h4mE0⇒λ1λ2=2.

The potential energy of a particle of mass ‘m’ varies with its position ‘x’ as The de-Broglie wavelengths of the particle when 0≤x≤1 and x>1 are λ1 and λ2 respectively. If the total energy of the particle is 2E0, then the value of λ1λ2 is

K.E. =2E0−E0=E0( for 0≤x≤1)⇒λ1=h2mE0 K.E. =2E0( for x>1)⇒ λ2=h4mE0⇒λ1λ2=2.

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The potential energy of a particle of mass ‘m’ varies with its position ‘x’ as
The de-Broglie wavelengths of the particle when $0 \leq x \leq 1$ and $x > 1$ are $λ_{1} a n d λ_{2}$ respectively. If the total energy of the particle is 2E₀, then the value of $(\frac{λ_{1}}{λ_{2}})$ is

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$2$

$1$

$\sqrt{2}$

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

answer is C.

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Detailed Solution

$\begin{array}{l} K.E. = 2 E_{0} - E_{0} = E_{0} (for 0 \leq x \leq 1) \Rightarrow λ_{1} = \frac{h}{\sqrt{2 m E_{0}}} \\ K.E. = 2 E_{0} (for x > 1) \Rightarrow λ_{2} = \frac{h}{\sqrt{4 m E_{0}}} \Rightarrow \frac{λ_{1}}{λ_{2}} = \sqrt{2} . \end{array}$