The graph between sqrt{E} and frac{1}{p} is ( | vedclass

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(C) The relationship between kinetic energy $E$ and momentum $p$ is given by $E = \frac{p^2}{2m}$,where $m$ is the mass of the particle.Taking the squ

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(C) The relationship between kinetic energy $E$ and momentum $p$ is given by $E = \frac{p^2}{2m}$,where $m$ is the mass of the particle.Taking the square root on both sides,we get $\sqrt{E} = \frac{p}{\sqrt{2m}}$.This can be rewritten as $\sqrt{E} = \frac{1}{\sqrt{2m}} \cdot p$.However,the question asks for the graph between $\sqrt{E}$ and $\frac{1}{p}$.From $E = \frac{p^2}{2m}$,we have $p = \sqrt{2mE}$.Therefore,$\frac{1}{p} = \frac{1}{\sqrt{2mE}} = \frac{1}{\sqrt{2m}} \cdot \frac{1}{\sqrt{E}}$.This implies $\sqrt{E} = \frac{1}{\sqrt{2m}} \cdot \frac{1}{(1/p)}$.Let $y = \sqrt{E}$ and $x = \frac{1}{p}$. Then $y = \frac{k}{x}$,where $k = \frac{1}{\sqrt{2m}}$ is a constant.This is the equation of a rectangular hyperbola. Thus,the graph between $\sqrt{E}$ and $\frac{1}{p}$ is a hyperbola.

The graph between $\sqrt{E}$ and $\frac{1}{p}$ is ($E$ = kinetic energy and $p$ = momentum).

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