The potential energy of a particle of mass m | vedclass

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(C) The de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $K$ is the kinetic energy of the particle.For the regio

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$\sqrt{2}$

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(C) The de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $K$ is the kinetic energy of the particle.For the region $0 \leqslant x \leqslant 1$,the potential energy is $U_1 = E_0$. Given the total energy $E = 2E_0$,the kinetic energy is $K_1 = E - U_1 = 2E_0 - E_0 = E_0$.Thus,$\lambda_1 = \frac{h}{\sqrt{2mE_0}}$.For the region $x > 1$,the potential energy is $U_2 = 0$. The kinetic energy is $K_2 = E - U_2 = 2E_0 - 0 = 2E_0$.Thus,$\lambda_2 = \frac{h}{\sqrt{2m(2E_0)}} = \frac{h}{\sqrt{4mE_0}}$.Taking the ratio $\frac{\lambda_1}{\lambda_2}$:$\frac{\lambda_1}{\lambda_2} = \frac{\frac{h}{\sqrt{2mE_0}}}{\frac{h}{\sqrt{4mE_0}}} = \sqrt{\frac{4mE_0}{2mE_0}} = \sqrt{2}$.

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