Dual Nature Of Light Question 343
Question: Electrons are accelerated through a potential difference V and protons are accelerated through a potential difference 4 V. The de-Broglie wavelengths are $ {\lambda_{e}} $ and $ {\lambda_{p}} $ for electrons and protons respectively. The ratio of $ \frac{{\lambda_{e}}}{{\lambda_{p}}} $ is given by: (given $ m_{e} $ is mass of electron and $ m_{p} $ is mass of proton)
Options:
A) $ \sqrt{\frac{m_{p}}{m_{e}}} $
B) $ \sqrt{\frac{m_{e}}{m_{p}}} $
C) $ \frac{1}{2}\sqrt{\frac{m_{e}}{m_{p}}} $
D) $ 2\sqrt{\frac{m_{e}}{m_{p}}} $
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Answer:
Correct Answer: D
Solution:
Energy in joule (E)= charge $ \times $ potential diff. in volt. $ E_{electron}=q_{e}V \text{ and }E_{proton}=q_{p}V $
De-Broglie wavelength $ {\lambda_{e}}=\frac{h}{\sqrt{2m_{e}eV}}\text{and }{\lambda_{p}}=\frac{h}{\sqrt{2m_{p}eV}} $
$ (\therefore q_{e}=q_{p}) $
$ \therefore \frac{{\lambda_{e}}}{{\lambda_{p}}}=\frac{\frac{h}{\sqrt{2m_{e}eV}}}{\frac{h}{\sqrt{2m_{p}eV}}}=\sqrt{\frac{2m_{p}eV}{2m_{e}eV}}=\sqrt{\frac{m_{p}}{m_{e}}} $
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