Em Waves Question 72
Question: The potential energy of a particle of mass m is given by $ U(x)= \begin{matrix} E _{0}; \ 0; \ \end{matrix}\begin{matrix} 0\le x\le 1 \ x>1 \ \end{matrix} . $ $ {\lambda _{1}} $ and $ {\lambda _{2}} $ are the de Broglie wavelengths of the particle, when $ 0\le x\le 1 $ and $ x>1 $ respectively. If the total energy of particle is 2 $ E _{0} $ , the ratio $ \frac{{\lambda _{0}}}{{\lambda _{2}}} $ will be
Options:
A) 2
B) 1
C) $ \sqrt{2} $
D) $ \frac{1}{\sqrt{2}} $
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Answer:
Correct Answer: C
Solution:
[c] $ K.E.=2E _{0}-E _{0}=(for0\le x\le 1) $
$ \Rightarrow {\lambda _{1}}=\frac{h}{\sqrt{2mE _{0}}} $
$ K.E.=2E _{0}(for,x>1) $
$ \Rightarrow {\lambda _{2}}=\frac{h}{\sqrt{4mE _{0}}}\Rightarrow \frac{{\lambda _{1}}}{{\lambda _{2}}}=\sqrt{2.} $
BETA
