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Potential energy of a particle varies as a function of $x$ and is given as $ U(x)$ \[ \begin {bmatrix} E_{\circ} & 0 \leq x \leq 1 \\ 0 & x > 1 \end {bmatrix} \] The de-Broglie wavelengths of the particle in the region $ 0 \leq x \leq1$ and $x>1$ are $ \lambda_1$ and $ \lambda_2$ respectively. If the total energy of the particle is $2E_{\circ}$, the ratio $ \large\frac{\lambda_1}{\lambda_2}$is

$\begin {array} {1 1} (a)\;\large\frac{1}{\sqrt 2} & \quad (b)\; \sqrt 2 \\ (c)\;2 & \quad (d)\;\large\frac{1}{2} \end {array}$

 

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Ans : (b)
Total energy = Kinetic energy + Potential energy
For the region $ 0 \leq x \leq 1,$
Kinetic energy = Total energy – Potential energy
$= 2E_{\circ} – E_{\circ} = E_{\circ}$
So, $ \lambda_1 = \large\frac{h}{ \sqrt {2m(E_{\circ})}}$…………..(i)
For the region $x>1$,
Kinetic energy = Total energy – Potential energy
$= 2E_{\circ} – 0 =2 E_{\circ}$
So, $ \lambda_2 = \large\frac{h}{ \sqrt{2m(2E_{\circ})}}$…………….(ii)
From (i) & (ii),
$ \large\frac{ \lambda_1}{ \lambda_2} = \sqrt 2$

 

answered Jan 1, 2014 by thanvigandhi_1
edited Mar 13, 2014 by thanvigandhi_1
 

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