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  • The potential energy of a particle varies as <img alt="\mathrm{ U(x)=E_0 \ \ 0 \leq x \leq 1 }" src="/latex-image/?%5Cmathrm%7B%20U%28x%29%3DE_0%20%5C%20%5C%200%20%5Cleq%20x%20%5Cleq%201%

The potential energy of a particle varies as

\mathrm{ U(x)=E_0 \ \ 0 \leq x \leq 1 }
\mathrm{ =0 \ \ \ x>1}
For \mathrm{0 \leq x \leq 1}. de-Broglie wavelength is \mathrm{\lambda_1 \: and \: for \: x>1} the de-Broglie wavelength is \mathrm{\lambda_2}. Total energy of the particle is \mathrm{2 E_0}. Then \mathrm{\frac{\lambda_1}{\lambda_2}} is:
 

Option: 1

2
 


Option: 2

2 \sqrt{2}


 


Option: 3

\sqrt{2}


Option: 4

1


Answers (1)

best_answer

For 0 \leq x \leq 1, \mathrm{PE}=E_0

\therefore Kinetic energy \mathrm{K_1=}Total energy -\mathrm{PE}

\mathrm{ =2 E_0-E_0=E_0 }
\mathrm{ \therefore \quad \lambda_1=\frac{h}{\sqrt{2 m\left(E_0\right)}} }      ......(1)

For \mathrm{ x>1, \mathrm{PE}=0 }

\mathrm{\therefore} Kinetic energy \mathrm{K_2=} Total energy \mathrm{=2 E_0}

\mathrm{\therefore \quad \lambda_2=\frac{h}{\sqrt{2 m\left(E_0\right)}}}             .........(2)

From equations. (1) and (2), we have

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

 

Posted by

shivangi.bhatnagar

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