The potential...

Question

The potential energy of a particle of mass m is given by
$U (x) = {\begin{matrix} E_{0}; & 0 \leq x \leq 1 0; & x > 1 \end{matrix}$
$λ_{1}$ and $λ_{2}$ 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 $3 E_{0}$ , the ratio $\frac{λ_{1}}{λ_{2}}$ will be

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\sqrt{2}

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\sqrt{\frac{3}{2}}

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Solution

The correct option is C $\sqrt{\frac{3}{2}}$
$K . E = 3 E_{0} - E_{0} = 2 E_{0} (f o r 0 \leq x \leq 1) \Rightarrow λ_{1} = \frac{h}{\sqrt{2 m_{2} E_{0}}}$
$K . E . = 3 E_{0} (f o r x > 1) \Rightarrow λ_{2} = \frac{h}{\sqrt{2 m_{3} E_{0}}} \Rightarrow \frac{λ_{1}}{λ_{2}} = \sqrt{\frac{3}{2}}$

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The correct option is C $\sqrt{\frac{3}{2}}$
$K . E = 3 E_{0} - E_{0} = 2 E_{0} (f o r 0 \leq x \leq 1) \Rightarrow λ_{1} = \frac{h}{\sqrt{2 m_{2} E_{0}}}$
$K . E . = 3 E_{0} (f o r x > 1) \Rightarrow λ_{2} = \frac{h}{\sqrt{2 m_{3} E_{0}}} \Rightarrow \frac{λ_{1}}{λ_{2}} = \sqrt{\frac{3}{2}}$