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The energy that should be added to an electron to reduce its de-Broglie wavelength from 1nm to 0.5 nm is

$\begin {array} {1 1} (a)\;4 \: times\: the\: initial\: energy & \quad (b)\;equal\: to\: initial \: energy \\ (c)\;twice\: the\: initial \: energy & \quad (d)\;thrice \: the\: initial\: energy \end {array}$

 

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Ans : (d)
$ \lambda = \large\frac{h}{ \sqrt{2mE}}$
$ \large\frac{ \lambda'}{ \lambda} = \large\frac{ \sqrt E}{E’}$
$ \Rightarrow \large\frac{E}{E’} = \bigg( \large\frac{\lambda’}{ \lambda} \bigg)^2 = \bigg( \large\frac{0.5}{1} \bigg)^2 =\large\frac{1}{4}$
$ \Rightarrow E’ = 4E$
So, Required energy = $E’ – E$
$= 4E - E = 3E$

 

answered Dec 31, 2013 by thanvigandhi_1
edited Mar 13, 2014 by thanvigandhi_1
 

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