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Standard XII

Physics

COE in SHM

The potential...

Question

The potential energy of a particle of mass m is given by $V (x) = {\begin{matrix} E_{0} 0 \end{matrix} \begin{matrix} 0 \leq x \leq 1 x > 1 \end{matrix}}$
$λ_{1}$ 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 ${2 E}_{0}$ , find $(λ_{1} / λ_{2})$ .

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Solution

Total energy , $E = T + V$ where T=Kinetic energy and V=potential energy.
When $0 \leq x \leq 1$ , $2 E_{0} = T + E_{0} \Rightarrow T = E_{0}$
de-Broglie wavelength, $λ_{1} = \frac{h}{\sqrt{2 m T}} = \frac{h}{\sqrt{2 m E_{0}}}$
When $x > 1, 2 E_{0} = T + 0$ or $T = 2 E_{0}$
now, $λ_{2} = \frac{h}{\sqrt{2 m T}} = \frac{h}{\sqrt{2 m 2 E_{0}}}$
Thus, $\frac{λ_{1}}{λ_{2}} = \sqrt{2}$

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The potential energy of a particle of mass m is given by V(x)={E000≤x≤1x>1}λ1 and λ2 are the de-Broglie wavelengths of the particle, when 0≤x≤1 and x>1 respectively.If the total energy of particle is 2E0, find (λ1/λ2).

Total energy , E=T+V where T=Kinetic energy and V=potential energy. When 0≤x≤1, 2E0=T+E0⇒T=E0de-Broglie wavelength, λ1=h√2mT=h√2mE0When x>1,2E0=T+0 or T=2E0now, λ2=h√2mT=h√2m2E0Thus, λ1λ2=√2