Dual Nature Of Light Question 111
Question: The potential energy of a particle of mass m is given by $ U(x)= E_{0};\ \ \ 0\le x\le 1 ,0\ ;\ \ \ x>1 . $ l1 and l2 are the de-Broglie wavelengths of the particle, when 0 £ x £ 1 and x > 1 respectively. If the total energy of particle is 2E0, the ratio $ \frac{{\lambda_{1}}}{{\lambda_{2}}} $ will be [Based on IIT-JEE (Mains) 2005]
Options:
A) 2
B) 1
C) $ \sqrt{2} $
D) $ \frac{1}{\sqrt{2}} $
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Answer:
Correct Answer: C
Solution:
K.E.= 2 E0? E0 = E0 (for 0 £ x £ 1)
Þ $ {\lambda_{1}}=\frac{h}{\sqrt{2m,E_{0}}} $ K.E. = 2 E0 (for x > 1)
Þ $ {\lambda_{2}}=\frac{h}{\sqrt{4m,E_{0}}} $
$ \Rightarrow \frac{{\lambda_{1}}}{{\lambda_{2}}}=\sqrt{2}. $
BETA
