A proton moving with a momentum p_{1} has a k | vedclass

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(D) For a proton,the total energy $E_{total} = E_k + mc^2 = \frac{1}{8}mc^2 + mc^2 = \frac{9}{8}mc^2$. Using the relativistic relation $E_{total}^2 =

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$3/4$

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(D) For a proton,the total energy $E_{total} = E_k + mc^2 = \frac{1}{8}mc^2 + mc^2 = \frac{9}{8}mc^2$. Using the relativistic relation $E_{total}^2 = (pc)^2 + (mc^2)^2$: $(\frac{9}{8}mc^2)^2 = (p_1c)^2 + (mc^2)^2$ $\frac{81}{64}m^2c^4 = p_1^2c^2 + m^2c^4$ $p_1^2c^2 = (\frac{81}{64} - 1)m^2c^4 = \frac{17}{64}m^2c^4$ $p_1 = \frac{\sqrt{17}}{8}mc$. For a photon,energy $E = p_2c = E_k = \frac{1}{8}mc^2$,so $p_2 = \frac{mc}{8}$. The ratio $\frac{p_1 - p_2}{p_1} = 1 - \frac{p_2}{p_1} = 1 - \frac{mc/8}{\sqrt{17}mc/8} = 1 - \frac{1}{\sqrt{17}}$. Note: If we assume non-relativistic kinetic energy $E_k = \frac{p_1^2}{2m} = \frac{1}{8}mc^2$,then $p_1^2 = \frac{1}{4}m^2c^2$,so $p_1 = \frac{mc}{2}$. Then $p_2 = \frac{E_k}{c} = \frac{mc}{8}$. The ratio $\frac{p_1 - p_2}{p_1} = \frac{mc/2 - mc/8}{mc/2} = \frac{3/8}{1/2} = \frac{3}{4}$.

$A$ proton moving with a momentum $p_{1}$ has a kinetic energy $1/8$th of its rest mass-energy. Another light photon having energy equal to the kinetic energy of the proton possesses a momentum $p_{2}$. Then,the ratio $\frac{p_{1}-p_{2}}{p_{1}}$ is equal to

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