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When the momentum of a proton is changed by an amount $p_{\circ}$, the corresponding change in de- Broglie wavelength is found to be $0.25 \%$. Then, the original momentum of the proton was

$\begin {array} {1 1} (a)\;p_{\circ} & \quad (b)\;100\: p_{\circ} \\ (c)\;400\: p_{\circ} & \quad (d)\;4\: p_{\circ} \end {array}$


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Ans : (c)
$ \lambda\: \sim\: \large\frac{1}{p}$
$ \Rightarrow \large\frac{\Delta p}{p} = -\large\frac{ \Delta \lambda}{ \lambda}$
$ \Rightarrow |\large\frac{ \Delta p}{p} | = | \large\frac{ \Delta \lambda}{ \lambda}|$
$\Rightarrow \large\frac{p_{\circ}}{p} = \large\frac{0.25}{100} = \large\frac{1}{400}$
$ \Rightarrow p = 400p_{\circ}$


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

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