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  • The potential energy of a particle varies as  <img alt="\begin{aligned} U(x) & =E_0 \quad \text { for } 0 \leq x \leq 1 \\ & =0, \quad x>1 . \end{aligned}" src="https://en

The potential energy of a particle varies as 

\begin{aligned} U(x) & =E_0 \quad \text { for } 0 \leq x \leq 1 \\ & =0, \quad x>1 . \end{aligned}
For 0 \leq x \leq 1, debroglie wavelength is \lambda_1 and for x>1 the debroglie wavelength is \lambda_2. Total energy of the particle is 2 E_0. Then value of \frac{\lambda_1}{\lambda_2} is ?

Option: 1

1


Option: 2

2


Option: 3

\sqrt{2}


Option: 4

\frac {1} {\sqrt{2}}


Answers (1)

best_answer

For 0 \leq x \leq 1, PE. =E_0
kinetic energy  k_1= Total energy -P \cdot E.
=2 E_0-E_0=E_0 \text {. }
\Rightarrow \lambda_1=\frac{h}{\sqrt{2 m E_0}}
For x>1, P E=0
\therefore kinetic energy  k_2 = Total energy = 2 E_0
\lambda_2=\frac{h}{\sqrt{4 m E_0}} \\

\Rightarrow \frac{\lambda_1}{\lambda_2}=\sqrt{2}

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mansi

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