1
IIT-JEE 2005 Mains
MCQ (Single Correct Answer)
+3
-1

The potential energy of a particle of mass m is given by

$$\mathrm{U}(x)=\left\{\begin{array}{cc}\mathrm{E}_{0} & 0 \leq x \leq 1 \\ 0 & x>1\end{array}\right.$$

$$\lambda_{1}$$ and $$\lambda_{2}$$ 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 $$2 \mathrm{E}_{0}$$, find $$\frac{\lambda_{1}}{\lambda_{2}}$$.

A
2
B
$$\sqrt2$$
C
$$\sqrt3$$
D
3

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