The following plot represents the de-Broglie | vedclass

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(C) The de-Broglie wavelength is given by the formula: $\lambda = \frac{h}{\sqrt{2m(K.E.)}} = \frac{h}{\sqrt{2m}} 	imes \frac{1}{\sqrt{K.E.}}$Compari

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$m_A > m_B$

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(C) The de-Broglie wavelength is given by the formula: $\lambda = \frac{h}{\sqrt{2m(K.E.)}} = \frac{h}{\sqrt{2m}} 	imes \frac{1}{\sqrt{K.E.}}$Comparing this with the equation of a straight line $y = mx$,where $y = \lambda$ and $x = \frac{1}{\sqrt{K.E.}}$,the slope is given by $	ext{slope} = \frac{h}{\sqrt{2m}}$.Since the slope is inversely proportional to the square root of the mass $(	ext{slope} \propto \frac{1}{\sqrt{m}})$,a larger slope indicates a smaller mass.From the graph,the slope of line $B$ is greater than the slope of line $A$ $(	ext{slope}_B > 	ext{slope}_A)$.Therefore,$m_B  m_B$.

The following plot represents the de-Broglie wavelength $(\lambda)$ as a function of $\frac{1}{\sqrt{K.E.}}$ for two particles $A$ and $B$. Identify the correct relation between their masses ($m_A$ and $m_B$).

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