The potential energy of a particle of mass m | vedclass

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(C) The total energy $E = 2E_0$. The kinetic energy is given by $K.E. = E - U(x)$.For $0 \le x \le 1$,$U(x) = E_0$,so $K.E._1 = 2E_0 - E_0 = E_0$.The

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$\sqrt{2}$

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(C) The total energy $E = 2E_0$. The kinetic energy is given by $K.E. = E - U(x)$.For $0 \le x \le 1$,$U(x) = E_0$,so $K.E._1 = 2E_0 - E_0 = E_0$.The de-Broglie wavelength is $\lambda_1 = \frac{h}{\sqrt{2m K.E._1}} = \frac{h}{\sqrt{2m E_0}}$.For $x > 1$,$U(x) = 0$,so $K.E._2 = 2E_0 - 0 = 2E_0$.The de-Broglie wavelength is $\lambda_2 = \frac{h}{\sqrt{2m K.E._2}} = \frac{h}{\sqrt{2m(2E_0)}} = \frac{h}{\sqrt{4m E_0}}$.Taking the ratio $\frac{\lambda_1}{\lambda_2} = \frac{h / \sqrt{2m E_0}}{h / \sqrt{4m E_0}} = \sqrt{\frac{4m E_0}{2m E_0}} = \sqrt{2}$.

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