The potential energy of a particle of mass m | vedclass

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(B) The total energy $E$ of the particle is given as $2E_0$.For the region $0 \le x \le 1$,the potential energy is $U(x) = E_0$. The kinetic energy $K

Vedclass · Accepted Answer

$2$

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(B) The total energy $E$ of the particle is given as $2E_0$.For the region $0 \le x \le 1$,the potential energy is $U(x) = E_0$. The kinetic energy $K_1$ is $E - U = 2E_0 - E_0 = E_0$.The de-Broglie wavelength is $\lambda_1 = \frac{h}{\sqrt{2mK_1}} = \frac{h}{\sqrt{2mE_0}}$.For the region $x > 1$,the potential energy is $U(x) = 0$. The kinetic energy $K_2$ is $E - U = 2E_0 - 0 = 2E_0$.The de-Broglie wavelength is $\lambda_2 = \frac{h}{\sqrt{2mK_2}} = \frac{h}{\sqrt{2m(2E_0)}} = \frac{h}{\sqrt{4mE_0}}$.Taking the ratio: $\frac{\lambda_1}{\lambda_2} = \frac{\frac{h}{\sqrt{2mE_0}}}{\frac{h}{\sqrt{4mE_0}}} = \sqrt{\frac{4mE_0}{2mE_0}} = \sqrt{2}$.Therefore,$(\lambda_1/\lambda_2)^2 = (\sqrt{2})^2 = 2$.

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