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An electron is accelerated through a potential difference of V volt. It has a wavelength $ \lambda$ associated with it. Through what potential difference must an electron be accelerated so that its de-Broglie wavelength is the same as that of a proton? Take mass of a proton to be 1837 times larger than the mass of electron.

$\begin {array} {1 1} (a)\;V\: volt & \quad (b)\;1837\: V\: volt \\ (c)\;\large\frac{V}{1837} volt & \quad (d)\;\sqrt{1837} V \: volt \end {array}$

 

1 Answer

Ans : (c)
$ \lambda = \large\frac{h}{ \sqrt{2meV}}$
Given, mV = constant; m for a proton is larger by a factor of 1837;
V is therefore smaller by a factor of 1837.

 

answered Dec 27, 2013 by thanvigandhi_1
edited Mar 13, 2014 by thanvigandhi_1
 

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