Dual Nature Of Light Question 134
Question: What is the de-Broglie wavelength of the $ \alpha - $ particle accelerated through a potential difference V [RPMT 1996]
Options:
A) $ \frac{0.287}{\sqrt{V}} $ Å
B) $ \frac{12.27}{\sqrt{V}} $ Å
C) $ \frac{0.101}{\sqrt{V}} $ Å
D) $ \frac{0.202}{\sqrt{V}} $ Å
Show Answer
Answer:
Correct Answer: C
Solution:
$ \lambda =\frac{h}{\sqrt{2mE}}=\frac{h}{\sqrt{2{m_{\alpha }}{Q_{\alpha }}V}} $ On putting $ {Q_{\alpha }}=2\times 1.6\times {{10}^{-19}}C $ $ {m_{\alpha }}=4m_{p}=4\times 1.67\times {{10}^{-27}}kg $ Þ $ \lambda =\frac{0.101}{\sqrt{V}}{\AA} $
BETA
